User:Dcljr/Series

Complete rewrite of Calculus/Series…

Note that "@@" marks things needing more attention before I post this in the main namespace.

Infinite series

A series is the sum of a sequence of terms. An infinite series is the sum of an infinite number of terms (the actual sum of the series need not be infinite, as we will see below).

While there is usually some pattern in the terms being added, technically a series can be the sum of any sequence of terms:

$1+2+\pi +\cos {\tfrac {\pi }{12}}+e+\ldots$

The problem with the above series is, it is not at all clear what the next term of the series would be, nor any subsequent ones. To say anything useful about the series, there needs to be a clear pattern. Two such patterns seen in sequences that also appear in the study of series are called arithmetic and geometric.

An arithmetic series is the sum of a sequence of terms having a common difference (i.e., the difference between consecutive terms is always the same). For example,

$1+4+7+10+13+\dots$

is an arithmetic series with common difference 3, since $a_{2}-a_{1}=3$, $a_{3}-a_{2}=3$, and so forth. Unlike the first series above, this series has a clear pattern. It is obvious that the next term of the series is 16, and then 19, and so on.

A geometric series is the sum of a sequence of terms having a common ratio (i.e., the ratio between consecutive terms is always the same). For example,

$1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\ldots$

is a geometric series with common ratio ${\tfrac {1}{2}}$, since $a_{2}/a_{1}={\tfrac {1}{2}}$, $a_{3}/a_{2}={\tfrac {1}{2}}$, and so forth. As before, there is a definite pattern, and the next term of the series is clearly ${\tfrac {1}{32}}$.

Summation notation

Summation notation provides a compact way of writing an infinite series. The arithmetic series above can be written

$\sum _{n=1}^{\infty }[1+3(n-1)]$

or, more simply

$\sum _{n=1}^{\infty }(3n-2).$

The symbol on the left is a Sigma (the Greek letter for capital "S", standing for "Sum"); it is also known as a summation symbol. Along with the expressions above and below the Sigma, it is read:

"The sum, as $n$ goes from 1 to infinity, of..."

The algebraic expression on the right is the nth term of the series, and is often generically called "$a_{n}$", as is also done with sequences. This means that the sum could have been written a bit more verbosely (and perhaps more pedantically) as

$\sum _{n=1}^{\infty }a_{n}{\mbox{, where }}a_{n}=3n-2,$

a form that more explicitly reflects the idea that a series is the sum of a sequence of terms. (The expression $a_{n}=3n-2$ by itself defines the sequence of terms, but note that it is a bit ambiguous since it's not clear without additional context whether the sequence starts at $n=1$ or $n=0$, or indeed any other value. Incidentally, the first summation formula above, which might have seemed a bit strange at first, came from the formula for the nth term of an arithmetic sequence, which you probably first learned in algebra class: $a_{n}=a_{1}+d(n-1)$.)

To verify that this summation formula represents the same sum as in the previous section, we simply evaluate the algebraic expression ($a_{n}$) for the first value of $n$ (in this case, 1), then the next value (2), and so forth:

$\sum _{n=1}^{\infty }(3n-2)=[3(1)-2]+[3(2)-2]+[3(3)-2]+\ldots =1+4+7+\ldots$

As for the geometric series, it should be easy to verify the representation

$\sum _{n=1}^{\infty }{\frac {1}{2^{n-1}}}$

or, equivalently

$\sum _{n=1}^{\infty }\left({\frac {1}{2}}\right)^{n-1}.$

Non-uniqueness of summation notation

Note that the representation of a series in summation notation is not unique, even ignoring simple algebraic rewrites. All of the sums seen so far have started at $n=1$, as is usually the convention adopted for sequences, but we could have chosen to start the sums at any other value of $n$. In fact, it is quite common to start infinite series at $n=0$. If we follow that convention, the arithmetic series would be written

$\sum _{n=0}^{\infty }(3n+1)$

and the geometric series would be

$\sum _{n=0}^{\infty }\left({\frac {1}{2}}\right)^{n}.$

Again, these expressions should be easy to verify.

Furthermore, as with functions, the variable used in the sum (called the index variable) is completely arbitrary. Thus, the following sums also represent the same series (respectively) as those above:

$\sum _{k=0}^{\infty }(3k+1)$

$\sum _{i=0}^{\infty }\left({\frac {1}{2}}\right)^{i}$

As a result of this fact, any series can be rewritten by simply changing the starting value of the series and compensating for this change in the formula for the nth term. To be specific, consider the first series above in this section. Using the substitution $n=m-2$, we get the series

$\sum _{m-2=0}^{\infty }(3[m-2]+1)=\sum _{m=2}^{\infty }(3m-5).$

But then the variable $m$ can simply be changed back to $n$ to get

$\sum _{n=2}^{\infty }(3n-5).$

Notice how this new series can be seen as resulting from "bumping up" the starting value of the original series by 2 and "bumping down" each $n$ in the $a_{n}$ formula by the same amount (that is, replacing each $n$ by $n-2$ and then simplifying). This kind of change to a series is sometimes necessary to verify a given series formula (say, when looking up the answer to an exercise in a textbook).

Series versus sequence

As mentioned above, there is an important difference between an infinite series, which is the sum of an infinite number of terms, and the sequence formed by those same terms.

At the risk of being repetitive, our arithmetic series is

$\sum _{n=1}^{\infty }(3n-2)=1+4+7+10+\ldots ,$

but the corresponding sequence of terms is

$\{a_{n}\}_{n=1}^{\infty }=\{3n-2\}_{n=1}^{\infty }=1,4,7,10,\ldots .$

It might seem silly at this point to make such a big deal out of this distinction, but it will be very important going forward not to confuse these two concepts.

Sequence of partial sums

One special kind of sequence that is very useful to consider when studying series is the sequence of partial sums, defined in the following way (assuming a starting value of 1 for the index variable) for any positive integer $n$:

$s_{n}=\sum _{k=1}^{n}a_{k}=a_{1}+a_{2}+a_{3}+\ldots +a_{n}$

For our arithmetic series, the sequence of partial sums is:

$s_{1}=a_{1}=1$

$s_{2}=a_{1}+a_{2}=1+4=5$

$s_{3}=a_{1}+a_{2}+a_{3}=1+4+7=12$

$s_{4}=a_{1}+a_{2}+a_{3}+a_{4}=1+4+7+10=22$

$\ldots$

Thus

$\left\{s_{n}\right\}=1,5,12,22,\ldots .$

Note how this sequence of partial sums is very different from the sequence of terms discussed in the last section!

Just as the series itself can be written more compactly in summation notation by finding the pattern in its terms, the sequence of partial sums for a series can (sometimes) also be written compactly as a function of $n$.

It should be easy to check that the following expression accurately represents the (first few terms of the) sequence of partial sums shown above:

$s_{n}={\frac {3n^{2}-n}{2}}$

To see where this expression came from, we first need to review some properties of sums that you probably remember from arithmetic and algebra, but might not be familiar with in summation notation.

Some properties of finite sums

Constant factors can be factored out of (or multiplied into) finite sums: $\sum _{k=1}^{n}c\,a_{k}=c\sum _{k=1}^{n}a_{k}$ Finite sums can be added and subtracted, as long as they cover the same range of values of the index variable: $\sum _{k=1}^{n}a_{k}+\sum _{k=1}^{n}b_{k}=\sum _{k=1}^{n}(a_{k}+b_{k})$ $\sum _{k=1}^{n}a_{k}-\sum _{k=1}^{n}b_{k}=\sum _{k=1}^{n}(a_{k}-b_{k})$ Putting these two ideas together, one may show the following property for arbitrary linear combinations: $\sum _{k=1}^{n}(c\,a_{k}+d\,b_{k})=c\sum _{k=1}^{n}a_{k}+d\sum _{k=1}^{n}b_{k}$ Furthermore, there are special formulas for sums of certain simple expressions: $\sum _{k=1}^{n}1=n$ $\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}$ $\sum _{k=1}^{n}r^{k}={\frac {r(1-r^{n})}{1-r}}{\mbox{ if }}r\neq 1$

The first special sum formula is obvious, since it represents the sum of n copies of the number 1. You might recognize the second and third formulas from your intermediate algebra or precalculus class.

With these facts, one may derive the expression given above for the sequence of partial sums of our arithmetic sequence.

${\begin{aligned}\sum _{k=1}^{n}(3k-2)&=\left(\sum _{k=1}^{n}3k\right)-\left(\sum _{k=1}^{n}2\right)\\&=3\left(\sum _{k=1}^{n}k\right)-2\left(\sum _{k=1}^{n}1\right)\\&=3\left[{\frac {n(n+1)}{2}}\right]-2(n)\\&={\frac {3n^{2}+3n}{2}}-{\frac {4n}{2}}\\s_{n}&={\frac {3n^{2}-n}{2}}\end{aligned}}$

Similarly, one may derive the sequence of partial sums for the geometric series:

${\begin{aligned}\sum _{k=1}^{n}({\tfrac {1}{2}})^{k-1}&=\sum _{k=1}^{n}({\tfrac {1}{2}})^{k}({\tfrac {1}{2}})^{-1}\\&=\sum _{k=1}^{n}({\tfrac {1}{2}})^{k}(2)\\&=2\sum _{k=1}^{n}({\tfrac {1}{2}})^{k}\\&=2\left({\frac {{\tfrac {1}{2}}[1-({\tfrac {1}{2}})^{n}]}{1-{\tfrac {1}{2}}}}\right)\\s_{n}&=2[1-({\tfrac {1}{2}})^{n}]\end{aligned}}$

Finding a series from its partial sums

Using the fact that

$s_{n}=a_{1}+a_{2}+\ldots +a_{n-1}+a_{n}=s_{n-1}+a_{n},$

we see that

$a_{n}=s_{n}-s_{n-1}.$

This provides a way of "recovering" the original series from its sequence of partial sums.

For our arithmetic series:

${\begin{aligned}a_{n}&=s_{n}-s_{n-1}\\&=\left[{\frac {3n^{2}-n}{2}}\right]-\left[{\frac {3(n-1)^{2}-(n-1)}{2}}\right]\\&=\left[{\frac {3n^{2}-n}{2}}\right]-\left[{\frac {3(n^{2}-2n+1)-(n-1)}{2}}\right]\\&=\left[{\frac {3n^{2}-n}{2}}\right]-\left[{\frac {3n^{2}-6n+3-n+1}{2}}\right]\\&={\frac {3n^{2}-n-3n^{2}+6n-3+n-1}{2}}\\&={\frac {6n-4}{2}}\\a_{n}&=3n-2\end{aligned}}$

Sum of an infinite series

Now we can finally formally define the sum of an infinite series as the limit of its sequence of partial sums:

$\sum _{n=1}^{\infty }a_{n}=\lim _{n\to \infty }s_{n}$

If the limit converges to a real number, say $s$, then the infinite series is said to converge to the sum $s$, or to be convergent with sum $s$. If the limit diverges (including the cases where the limit is infinity or negative infinity), then the series is said to diverge or to be divergent in the same way, and its sum is said to not exist (or to be infinity or negative infinity, as appropriate). Note that one does not describe the series itself as "not existing" when it diverges, only its sum.

Consider again the arithmetic and geometric series we have been discussing up to this point. It is obvious by simply looking at the original arithmetic series

$1+4+7+10+13+\dots$

that it does not have a finite sum. The terms being added are themselves getting larger and larger without bound, so the sum is getting larger and larger without bound.

The sequence of partial sums given above formalizes this idea. In particular, because

$\sum _{n=1}^{\infty }(3n-2)=\lim _{n\to \infty }{\frac {3n^{2}-n}{2}}=\infty ,$

the arithmetic series diverges to infinity.

Now consider the original geometric series:

$1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\ldots$

The terms here are getting smaller and smaller, and indeed are approaching zero. Since adding zero to something doesn't change its value, it seems reasonable to suspect that the sum might be a fixed, finite number. What does the sequence of partial sums reveal?

$\sum _{n=1}^{\infty }({\tfrac {1}{2}})^{n-1}=\lim _{n\to \infty }2[1-({\tfrac {1}{2}})^{n}]=2(1-0)=2$

So the geometric series does, in fact, converge to the finite sum 2.

It seems obvious that any series whose terms "blow up" to infinity (like our arithmetic series) will diverge, but does every series whose terms shrink to zero (like our geometric series) converge to a finite sum? It turns out the answer to that question is no.

Harmonic series

There is a very important series whose sequence of terms goes to zero and yet the series diverges because the sequence of partial sums diverges to infinity. It is called the harmonic series:

$\sum _{n=1}^{\infty }{\frac {1}{n}}$

Obviously the terms of this series go to zero:

$\lim _{n\to \infty }{\frac {1}{n}}=0$

But what about the sequence of partial sums? For convenience sake, we consider not the nth partial sum but the $2^{n}$th partial sum, then group the terms in a clever way, and find a lower bound for each group of terms:

${\begin{matrix}\displaystyle {\sum _{k=1}^{2^{n}}{\frac {1}{k}}}&=1+{}&\underbrace {\frac {1}{2}} _{{\text{group }}1}&+&\underbrace {{\frac {1}{3}}+{\frac {1}{4}}} _{{\text{group }}2}&+&\underbrace {{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}} _{{\text{group }}3}&{}+\ldots +{}&\underbrace {\sum _{k=2^{n-1}+1}^{2^{n}}{\frac {1}{k}}} _{{\text{group }}n}\\&>1+{}&{\frac {1}{2}}&+&{\frac {1}{4}}(2)&+&{\frac {1}{8}}(4)&{}+\ldots +{}&{\frac {1}{2^{n}}}(2^{n-1})\\\\&=1+{}&{\frac {1}{2}}&+&{\frac {1}{2}}&+&{\frac {1}{2}}&{}+\ldots +{}&{\frac {1}{2}}\end{matrix}}$

There are n terms equal to ${\tfrac {1}{2}}$ in the final sum, thus

$\sum _{k=1}^{2^{n}}{\frac {1}{k}}>1+\sum _{k=1}^{n}{\frac {1}{2}}=1+{\frac {n}{2}}.$

The limit of this final expression, as $n\to \infty$, is infinity. This means the sequence of partial sums diverges. (Technically, we have only shown that a "subsequence" of the original sequence of partial sums diverges, but it turns out that this is sufficient to prove that the original sequence diverges.)

Therefore, the harmonic series diverges even though its terms shrink to zero. Many other series share this property, some of which will be discussed below.

"Eventually"

When considering whether a given series is convergent or divergent, it is often useful to "ignore" the first several terms of the series. For example, as we have seen, the series

$1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\ldots$

is geometric and converges to 2, but the series

$-1+3+1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\ldots$

is not geometric, since the first two terms don't fit the pattern of the rest of the series. But since the series behaves like a geometric series from the third term onward, one might call the series "eventually" geometric: it didn't start out that way, but eventually it settled down and behaved like a geometric series. We will use this idea of a series "eventually" having a certain property many times in the discussion that follows.

Clearly if the first (geometric) series above converges, the second (eventually geometric) series will also converge. In fact, by considering separately the sum of the first two terms and the sum of the rest of the terms, we can deduce that the sum of the second series is

$-1+3+\sum _{n=1}^{\infty }{\frac {1}{2^{n-1}}}=(-1+3)+(2)=4.$

Similarly, the series

${\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\ldots$

converges in the same way as the original geometric series, but again the new (this time shorter) series converges to a different sum (the original sum minus the first two terms):

$\sum _{n=3}^{\infty }{\frac {1}{2^{n-1}}}={\biggl (}\sum _{n=1}^{\infty }{\frac {1}{2^{n-1}}}{\biggr )}-{\biggl (}\sum _{n=1}^{2}{\frac {1}{2^{n-1}}}{\biggr )}=(2)-(1+{\tfrac {1}{2}})={\tfrac {1}{2}}$

So, note that whether a series converges doesn't depend on what's happening at the beginning of the series, but the sum of the series definitely does!

Series convergence and divergence

Now that we've covered enough background about series, we can start to systematically investigate the convergence and divergence of many different kinds of infinite series. In some cases we will be able to find a formula for the partial sums, and thus the sum of the series, but in most cases we will not.

Necessary condition for convergence

Note that we have already considered one convergent series whose terms shrink to zero (the geometric series), one divergent series whose terms shrink to zero (the harmonic series), and one divergent series whose terms do not shrink to zero (the arithmetic series). What about the remaining case: a convergent series whose terms do not shrink to zero? Turns out, that's not possible.

Theorem: A necessary condition for series convergence If the infinite series $\sum _{n=c}^{\infty }a_{n}$ converges (where $c$ is any integer), then $\lim _{n\to \infty }a_{n}=0$.

Proof: to come ∎

So, every convergent series has a sequence of terms that converges to zero. But, as we saw with the harmonic series, just because a series has terms that shrink to zero, that doesn't mean that the series converges. Therefore, having terms that go to zero is a necessary condition that a convergent series must satisfy, but not a sufficient condition that allows us to conclude that a given series converges. (Put more bluntly: Just because the terms go to zero doesn't mean the series converges!)

So what is a sufficient condition for series convergence? Well, there are many such conditions, several of which we will discuss below. All of them, however, will be saying essentially the same thing: that the terms of the series are going to zero "fast enough" that it allows the series to converge.

Sufficient condition for divergence

Recall from basic logic that the statement "If A then B" is equivalent to "If not B then not A". Applying this idea to the previous theorem gives a way of determining that an infinite series diverges (the so-called divergence test, which we will see again below).

Theorem: A sufficient condition for series divergence If it is not true that $\lim _{n\to \infty }a_{n}=0$ (including the case in which the limit does not exist), then the infinite series $\sum _{n=c}^{\infty }a_{n}$ (where $c$ is any integer) diverges.

Proof: This is an immediate consequence of the last theorem. The argument is a simple proof by contradiction: Let $\lim _{n\to \infty }a_{n}$ either not equal 0 or not exist. Assume that $\sum _{n=c}^{\infty }a_{n}$ converges (for some integer $c$). Then by the theorem above, $\lim _{n\to \infty }a_{n}$ must equal 0. This is clearly a contradiction. Thus $\sum _{n=c}^{\infty }a_{n}$ must diverge. ∎

While this gives a very useful way of checking for divergence, it is not a very "powerful" method, since — as we've already said — there are many divergent infinite series whose terms do actually go to zero. Several more powerful methods for checking divergence will be given below.@@

Example

to come@@

Special forms

It is useful to distinguish between series with special forms, which all have specific formulas associated with them, and general-purpose tests that can be applied to a wide variety of different series.

Because the only arithmetic series that converges is the "trivial" series consisting of all zeros, we will not consider that type of infinite series any further.

Geometric series

A geometric series any series that can be written in either of the two equivalent forms $\sum _{n=1}^{\infty }a\,r^{n-1}{\mbox{ or }}\sum _{n=0}^{\infty }a\,r^{n},$ where $a$ is any nonzero real number and $r$ is any real number.

These two forms are the most common ones given in calculus textbooks, but note that both are special cases of the more general form

$\sum _{n=c}^{\infty }a\,r^{n-c},$

where, again, $a$ is any nonzero real number, $r$ any real number, and $c$ any integer.

Using any of the forms above, the constant $a$ is the first term of the series and the constant $r$ is the common ratio between successive terms; in particular,

$r=a_{n+1}-a_{n}$

for any $n$ for which both terms are defined.

The partial sums of such a geometric series are given by $s_{n}={\frac {a(1-r^{n})}{1-r}}.$ The series converges if and only if $|r|<1$, and in this case the sum of the series is $s={\frac {a}{1-r}}.$

Note that this last formula can be used with any $r\neq 1$, but it only is meaningful (and represents the sum of the series) when $|r|<1$, because otherwise the series doesn't even converge. In other words, be sure to check that the series converges before plugging into the formula for the sum.

As discussed in a previous section, the convergence or divergence of a series doesn't depend on where or how the series starts, so we could have defined a geometric series as starting at any $n$ value. But the formulas for the n^th partial sum and the sum of the series given above only work when the first term is $a$ and the common ratio is $r$. Consider the series

$\sum _{n=1}^{\infty }ar^{n}$

This is definitely a geometric series, but note that the first term is not $a$, it is $ar$, so the partial sums and the sum of the series will not be as stated above. In this case, a simple algebraic rewrite fixes the problem:

$\sum _{n=1}^{\infty }ar^{n}=\sum _{n=1}^{\infty }ar\,r^{n-1}=r\sum _{n=1}^{\infty }ar^{n-1}$

We see, therefore, that the partial sums and sum of the series are the ones given above multiplied by $r$.

To avoid having to rewrite an "obviously geometric" series just to get the same form as in the definition, the following general formulas can be used instead:

$s_{n}={\dfrac {\langle {\text{first term}}\rangle -\langle (n+1){\text{-th term}}\rangle }{1-\langle {\text{common ratio}}\rangle }}$ $s={\dfrac {\langle {\text{first term}}\rangle }{1-\langle {\text{common ratio}}\rangle }}$

Again, these formulas should only be used after one has verified that the series is actually geometric and, if one is looking for the sum, that the series actually converges.

Example: Consider the series $\sum _{n=2}^{\infty }{\frac {3^{n+1}}{5^{n-1}}}.$ It's pretty obvious that this series is geometric with common ratio ${\tfrac {3}{5}}$ (because when $n$ increases by 1, the numerator increases by a factor of 3 and the denominator by a factor of 5) and that the first term is ${\tfrac {27}{5}}$. Since $|{\tfrac {3}{5}}|<1$, the series converges to ${\frac {\tfrac {27}{5}}{1-{\tfrac {3}{5}}}}={\frac {27}{5-3}}={\frac {27}{2}}.$ To write the series in one of the definitional forms given above, notice that $\sum _{n=2}^{\infty }{\frac {3^{n+1}}{5^{n-1}}}=\sum _{n=2}^{\infty }9\left({\frac {3}{5}}\right)^{n-1}=\sum _{n=2}^{\infty }{\frac {27}{5}}\left({\frac {3}{5}}\right)^{n-2}=\sum _{n=0}^{\infty }{\frac {27}{5}}\left({\frac {3}{5}}\right)^{n}$ which implies the same first term and common ratio as previously stated.

Telescoping series

A telescoping series (or telescoping sum) is one that "expands" in such a way that most of its terms cancel away. Typically this will be a series of the form $\sum _{n=c}^{\infty }{\frac {d}{(n+a)(n+b)}}$ where $a$, $b$, and $c$ are all integers with $a<b$ (assuming no division by zero occurs for any of the terms) and $d$ any nonzero real number. The partial sums of such a series can be written in the form $\sum _{k=c}^{n}\left({\frac {A}{k+a}}+{\frac {B}{k+b}}\right)$ by way of a partial fractions decomposition.

By expanding this finite series, one will see that everything cancels except portions of its first $a-b$ terms (i.e., the terms corresponding to $k=c,c+1,\ldots ,c+(b-a)-1$) and last $a-b$ terms (for $k=n-(b-a)+1,\ldots ,n$). It should then be easy to find the limit of the partial sums and hence the sum of the series.

As we will see later,@@ any telescoping series of the particular form shown above must converge, but there are telescoping series of other forms which do not converge.

Example: Consider the series $\sum _{n=1}^{\infty }{\frac {1}{n^{2}+2n}}.$ This can be written as $\sum _{n=1}^{\infty }{\frac {1}{n(n+2)}}$ and so is telescoping with $a=0$, $b=2$, $c=1$, and $d=1$. The partial sums of the series are $s_{n}=\sum _{k=1}^{n}{\frac {1}{k(k+2)}}=\sum _{k=1}^{n}\left({\frac {1/2}{k}}-{\frac {1/2}{k+2}}\right)=\sum _{k=1}^{n}\left({\frac {1}{2k}}-{\frac {1}{2k+4}}\right)$ by a partial fractions decomposition. Expanding the sum, we get ${\begin{aligned}s_{n}=&\left({\frac {1}{2}}-{\frac {1}{6}}\right)+\left({\frac {1}{4}}-{\frac {1}{8}}\right)+\left({\frac {1}{6}}-{\frac {1}{10}}\right)\\&{}+\left({\frac {1}{8}}-{\frac {1}{12}}\right)+\left({\frac {1}{10}}-{\frac {1}{14}}\right)+\dots +\left({\frac {1}{2n-8}}-{\frac {1}{2n-4}}\right)+\left({\frac {1}{2n-6}}-{\frac {1}{2n-2}}\right)\\&{}+\left({\frac {1}{2n-4}}-{\frac {1}{2n}}\right)+\left({\frac {1}{2n-2}}-{\frac {1}{2n+2}}\right)+\left({\frac {1}{2n}}-{\frac {1}{2n+4}}\right).\end{aligned}}$ Notice that the fractions ${\tfrac {1}{6}}$, ${\tfrac {1}{8}}$, and ${\tfrac {1}{10}}$ immediately cancel, as do ${\tfrac {1}{2n-4}}$, ${\tfrac {1}{2n-2}}$, and ${\tfrac {1}{2n}}$. But by recognizing the pattern of cancelations, it should be clear that the fractions ${\tfrac {1}{12}}$ and ${\tfrac {1}{14}}$ also cancel at the beginning of the series (with later terms) and ${\tfrac {1}{2n-8}}$ and ${\tfrac {1}{2n-6}}$ also cancel at the end of the series (with earlier terms). Similarly, all the fractions in the omitted portion of the series ($\dots$) also cancel away. However, the fractions ${\tfrac {1}{2}}$ and ${\tfrac {1}{4}}$ at the beginning as well as ${\tfrac {1}{2n+2}}$ and ${\tfrac {1}{2n+4}}$ at the end do not cancel out. This is as we expect, since the fact that $b-a=2$ tells us that portions of the first two and last two terms will remain after canceling. In this way we see that $s_{n}={\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{2n+2}}-{\frac {1}{2n+4}}$ and so the sum of the series is $s=\lim _{n\to \infty }s_{n}=\lim _{n\to \infty }\left({\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{2n+2}}-{\frac {1}{2n+4}}\right)={\frac {1}{2}}+{\frac {1}{4}}-0-0={\frac {3}{4}}.$

p-series

A @@p-series is any series that can be written in the form

$\sum _{n=c}^{\infty }{\frac {a}{n^{p}}}$

for some positive integer $c$, nonzero real number $a$, and real number $p$. Such a series converges if and only if $p>1$.

The partial sums of p-series are too complicated to write a general formula for, so we won't try to do this. Nor will we consider the sum that such a series converges to. There are methods that can be used to find the sums of certain particular p-series, but this will have to wait until we discuss power series.

Note that the harmonic series discussed above is a p-series with $p=1$ (and $a=1$).

Example

Consider the series

$\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}.$

This series can be written as

$\sum _{n=1}^{\infty }{\frac {1}{n^{1/2}}}$

and so is a p-series with $p={\tfrac {1}{2}}$ (and $c=1$). Since $p<1$, the series diverges.

Alternating series

An alternating series is any series whose terms alternate in sign — that is, any series for which the product of any two consecutive terms is negative.

Equivalently, an alternating series is one that can be written in the form

$\sum _{n=c}^{\infty }(-1)^{n+d}\,b_{n},$

for some fixed integers $c$ and $d$, and sequence of positive terms $b_{n}$. (If the n^th term of the series is called $a_{n}$, as usual, then notice that $b_{n}=|a_{n}|$.)

As with geometric series, we have defined alternating series here in a slightly more general way than is typically done in calculus textbooks. Usually alternating series are either defined quite restrictively as

$\sum _{n=1}^{\infty }(-1)^{n+1}b_{n},$

or as being in one of the two forms

$\sum _{n=1}^{\infty }(-1)^{n}b_{n}{\mbox{ or }}\sum _{n=1}^{\infty }(-1)^{n-1}b_{n},$

in all cases the $b_{n}$ being some sequence of positive terms. It is easy to see that these other definitions are special cases of our formula above.

It should be obvious that we could not hope to write a formula for the partial sums of a general alternating series (besides, of course, the definition of partial sums given earlier), but, perhaps surprisingly, we can say when such a series converges.

An alternating series converges if its terms eventually decrease in magnitude to zero — that is, if

$\lim _{n\to \infty }|a_{n}|=0$

and

the sequence $\{|a_{n}|\}$ is eventually decreasing.

However, if either of these conditions are not satisfied, it does not mean that the alternating series must diverge.

Note that any geometric series with $r<0$ is alternating. If $|r|<1$, then the conditions for convergence of an alternating series will be satisfied. This can be proven in the general case,@@ but we will simply illustrate with an example.

Example

Consider the series

$\sum _{n=1}^{\infty }\left(-{\frac {1}{2}}\right)^{n-1}.$

This series can be written in the form

$\sum _{n=1}^{\infty }(-1)^{n-1}\left({\tfrac {1}{2}}\right)^{n-1},$

and so matches our definition of an alternating series ($c=1$, $d=-1$, and $b_{n}=({\tfrac {1}{2}})^{n-1}$).

It is obvious that

$\lim _{n\to \infty }|a_{n}|=\lim _{n\to \infty }\left({\frac {1}{2}}\right)^{n-1}=0$

and that

$|a_{n}|=\left({\frac {1}{2}}\right)^{n-1}$

is a decreasing sequence, so the series converges (as we knew it must, since it is geometric with $|r|<1$).

Example

Consider the series

$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}.$

This is an alternating version of the harmonic series. Since

$\lim _{n\to \infty }{\frac {1}{n}}=0$

and

$|a_{n}|={\frac {1}{n}}$

is a decreasing sequence of terms, the series converges.

Absolute and conditional convergence

At this point we have considered divergent series whose terms have no limit (the arithmetic series) and divergent series whose terms have a limit of zero (the harmonic series). But every convergent series must have terms that converge to zero. So, does this mean convergent series are "all the same"? Definitely not. There are two kinds of convergence that can be thought of as two "strengths" of convergence: absolute and conditional. The distinction is important because there are things you can do with absolutely convergent series that you cannot do with merely conditionally convergent ones. First, though, some definitions:

• A series $\textstyle \sum _{n=c}^{\infty }a_{n}$ is said to converge absolutely (or to be absolutely convergent) if $\textstyle \sum _{n=c}^{\infty }|a_{n}|$ converges.
• A series $\textstyle \sum _{n=c}^{\infty }a_{n}$ is said to converge conditionally (or to be conditionally convergent) if $\textstyle \sum _{n=c}^{\infty }|a_{n}|$ diverges but $\textstyle \sum _{n=c}^{\infty }a_{n}$ itself converges.

It should be obvious that this distinction only makes sense for series with a mixture of positive and negative terms. This includes, but is not limited to, alternating series.

Example

We have seen that the harmonic series

$\sum _{n=1}^{\infty }{\frac {1}{n}}$

is divergent but its alternating version

$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}$

is convergent. Since the first series may be formed by taking the absolute value of the terms in the second series, we see that the second series is conditionally convergent.

Example

Consider the series

$\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n^{2}}}.$

This is an alternating series whose terms decrease to zero in magnitude, so it converges. Furthermore, the series formed by the absolute value of the terms

$\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$

is a p-series with $p=2$, so it converges also. Therefore the original alternating series is absolutely convergent.

Properties of infinite series

There are several properties of infinite series that are direct consequences of the corresponding properties of finite sums:

Constants can be factored out of (or multiplied into) infinite series:

• For any real number $c$, $\textstyle \sum _{k=1}^{\infty }c\,a_{k}$ converges if $\textstyle \sum _{k=1}^{\infty }a_{k}$ converges, in which case $\textstyle \sum _{k=1}^{\infty }c\,a_{k}=c\sum _{k=1}^{\infty }a_{k}.$
• For any non-zero real number $c$, $\textstyle \sum _{k=1}^{\infty }c\,a_{k}$ diverges if $\textstyle \sum _{k=1}^{\infty }a_{k}$ diverges.

Note that $\textstyle \sum _{k=1}^{\infty }c\,a_{k}$ converges if $\textstyle \sum _{k=1}^{\infty }a_{k}$ diverges but $c=0$ (because it is a series whose terms are all zero).

Sums and differences of convergent infinite series are convergent:

• If $\textstyle \sum _{k=1}^{\infty }a_{k}$ and $\textstyle \sum _{k=1}^{\infty }b_{k}$ both converge, then $\textstyle \sum _{k=1}^{\infty }(a_{k}+b_{k})$ also converges and $\textstyle \sum _{k=1}^{\infty }(a_{k}+b_{k})=\sum _{k=1}^{\infty }a_{k}+\sum _{k=1}^{\infty }b_{k}.$
• If $\textstyle \sum _{k=1}^{\infty }a_{k}$ and $\textstyle \sum _{k=1}^{\infty }b_{k}$ both converge, then $\textstyle \sum _{k=1}^{\infty }(a_{k}-b_{k})$ also converges and $\textstyle \sum _{k=1}^{\infty }(a_{k}-b_{k})=\sum _{k=1}^{\infty }a_{k}-\sum _{k=1}^{\infty }b_{k}.$

Sums and differences of one convergent and one divergent series are divergent:

• If $\textstyle \sum _{k=1}^{\infty }a_{k}$ converges and $\textstyle \sum _{k=1}^{\infty }b_{k}$ diverges, then $\textstyle \sum _{k=1}^{\infty }(a_{k}+b_{k})$ and $\textstyle \sum _{k=1}^{\infty }(a_{k}-b_{k})$ both diverge.
• If $\textstyle \sum _{k=1}^{\infty }a_{k}$ diverges and $\textstyle \sum _{k=1}^{\infty }b_{k}$ converges, then $\textstyle \sum _{k=1}^{\infty }(a_{k}+b_{k})$ and $\textstyle \sum _{k=1}^{\infty }(a_{k}-b_{k})$ both diverge.

Sums and differences of two divergent series may be convergent or divergent (thus one cannot decide whether they converge or diverge without applying a specific convergence or divergence test):

• If $\textstyle \sum _{k=1}^{\infty }a_{k}$ and $\textstyle \sum _{k=1}^{\infty }b_{k}$ both diverge, then $\textstyle \sum _{k=1}^{\infty }(a_{k}+b_{k})$ and $\textstyle \sum _{k=1}^{\infty }(a_{k}-b_{k})$ may either (both) converge or (both) diverge.

More generally, linear combinations of convergent infinite series are convergent:

• If $\textstyle \sum _{k=1}^{\infty }a_{k}$ and $\textstyle \sum _{k=1}^{\infty }b_{k}$ both converge and $c$ and $d$ are any real numbers, then $\textstyle \sum _{k=1}^{\infty }(c\,a_{k}+d\,b_{k})$ also converges and $\textstyle \sum _{k=1}^{\infty }(c\,a_{k}+d\,b_{k})=c\sum _{k=1}^{\infty }a_{k}+d\sum _{k=1}^{\infty }b_{k}.$

Linear combinations of a mix of convergent and divergent series are (usually) divergent:

• If $\textstyle \sum _{k=1}^{\infty }a_{k}$ converges and $\textstyle \sum _{k=1}^{\infty }b_{k}$ diverges, and $c$ and $d$ are any real numbers with $d\neq 0$, then $\textstyle \sum _{k=1}^{\infty }(c\,a_{k}+d\,b_{k})$ diverges. (If $d=0$, then $\textstyle \sum _{k=1}^{\infty }(c\,a_{k}+d\,b_{k})$ converges.)

However, arbitrary linear combinations of divergent series are (usually) inconclusive with respect to their convergence without further testing:

• If $\textstyle \sum _{k=1}^{\infty }a_{k}$ and $\textstyle \sum _{k=1}^{\infty }b_{k}$ both diverge and $c$ and $d$ are non-zero real numbers, then $\textstyle \sum _{k=1}^{\infty }(c\,a_{k}+d\,b_{k})$ may either converge or diverge. (If exactly one of the constants $c$ and $d$ is zero, then the latter series diverges. If both $c=0$ and $d=0$, then the latter series converges.)

Special properties of absolutely convergent series

to come@@

Example

to come@@

Tests for convergence and divergence

In this section we state and illustrate several general-purpose tests for convergence and divergence of infinite series. These tests, along with the special forms listed above are sufficient to classify most infinite series that come up in calculus classes as convergent (absolutely or conditionally) or divergent. However, not every test will work on a given series (that's why there are so many of them), so understanding which test is appropriate for which kind of series is crucial.@@

Divergence test

We have already seen the first general-purpose test for infinite series in the section above about a sufficient condition for divergence. It never concludes convergence, only divergence, and is therefore called the Divergence Test (it's also known as the n^th-Term Test, or sometimes the Limit Test).

Theorem: Divergence test for infinite series If $\lim _{n\to \infty }a_{n}\neq 0$ or $\lim _{n\to \infty }a_{n}$ does not exist, then the infinite series $\sum _{n=c}^{\infty }a_{n}$ (where $c$ is any integer) diverges.

Proof: See the theorem under Sufficient condition for divergence. ∎

It is very important to remember that this test can only conclude divergence (hence the name); it cannot be used to conclude convergence. In particular, this means that if the limit of the terms is zero, then you don't conclude that the series is "not divergent" (since that means it is convergent). Instead, you say that the test is inconclusive and don't conclude anything about the series: some other test is then required to say whether the series converges or diverges.

Example

Example

Integral test

Example

Approximating the sum of a series with positive terms

Example

Comparison tests

Direct comparison test

Example

Example

Limit comparison test

Example

Picking a series for comparison

Example

Alternating series test

Example

Absolute and conditional convergence

Example

Approximating the sum of an alternating series

Example

Root test

Example

Ratio test

Example