# Real Analysis/Series

Frequently in analysis it is useful to consider the sum of infinitely many numbers. So for some sequence of numbers (an) we may want to consider expressions like

$\sum_{n=1}^\infty a_n$.

But the exact meaning of this may not be immediately clear. Intuitively we would like to say that the sum of infinitely many numbers should be the number we get close to after we have summed a large number of terms. We will use the notion of the limit of a sequence to make this precise. The standard terminology in the study of series sometimes has room for improvement, but we follow the standard terminology in this section.

## DefinitionEdit

We begin with a sequence (an) of the numbers we would like to sum.

Definition A series of real numbers is an infinite formal sum
$\sum_{n=1}^\infty a_n$
where each term an is a real number.

This definition deserves a few comments. The first is that it that no attempt is being made to define a formal sum. It is possible to define a formal sum simply as the sequence of terms, but this doesn't add any clarity to the discussion. The reason for allowing the series to be formal is merely a matter of convenience. It is frequently easier to refer to a series before it as been determined if there is any number that should represent the sum. This is very similar to the standard practice of saying $\textstyle \lim_{n\to\infty}a_n$ does not exists, we have only defined the meaning of the symbols $\textstyle \lim_{n\to\infty}a_n$ in the case when the limit does exist. We should instead say that the sequence an does not converge. However the meaning of the previous statement is perfectly clear.

Definition The n-th partial sum of a sequence an is defined to be the sum of the first n terms of (an), that is
$S_n=\sum_{i=1}^n a_i=a_1+a_2+\cdots+a_n$.
When the sequence an is being thought of as the terms of a series, then Sn is often called the n-th partial sum of the series.

Often several partial sums may appear in the same argument, so the partial sum is often written simply as $\sum_{i=1}^n a_i$ instead of Sn when we wish to avoid confusion.

Definition For a series $\textstyle \sum_{n=1}^\infty a_n$ we define the sum of the series to be the limit of the partial sums. That is, we define:
$\sum_{n=1}^{\infty} a_{n}=\lim_{N\to\infty} \sum_{n=1}^N a_n=\lim_{N\to\infty}S_N$.
If the limit exists we say the series converges, otherwise we say the series diverges.

It should be emphasized when a series diverges, we cannot interpret $\textstyle \sum_{n=1}^\infty a_n$ as a number, but only as a formal sum. On the other hand when the series does converge, it is often not known what number the series converges to, so this number is usually denoted by $\textstyle \sum_{n=1}^\infty a_n$. Again this is similar to writing $\textstyle \lim_{n\to\infty}a_n$ before the convergence of the sequence an is established. In practice the meaning of the symbols $\textstyle \sum_{n=1}^\infty a_n$ is clear from context.

Often it is convenient to sum series starting from some number other than n = 1, and start the series some other point like 0, 2, −10, etc. This hopefully should cause no confusion, the sum of the series is still defined as the limit of the partial sums. Often it is sometimes clear from context where the sum begins, in these cases it is not uncommon to leave the index out of the sigma notation, that is it is useful to write ∑an for $\textstyle \sum_{n=1}^\infty a_n$

## ExamplesEdit

• The notion of an infinite sum can be a little more subtle than it first appears. For example, consider the sequence an = (−1)n. Does the sum of the an converge or diverge? We could consider the partial sums are SN which are 1 if N is odd and 0 if N is even. So it seems the series diverges. On the other hand 1 − 1 + 1 − 1 + 1 − 1 + … = (1 − 1) + (1 − 1) + (1 − 1) + … = 0 + 0 + 0 + … = 0. Does our theory say the series diverges or does it converge to 0? The answer is it diverges, the fallacy in the previous sentence was in asserting that 1 − 1 + 1 − 1 + 1 − 1 + … = (1 − 1) + (1 − 1) + (1 − 1) + …. It is no longer true that for infinite series that we may sum in any order we choose, we would need to justify why we are allowed to group the +1's together with the -1's without changing the sum. We will investigate when it is possible to rearrange the elements of a series and still get the same sum.
• Perhaps the most familiar examples of series come from decimal expansions or real numbers. While we have not given a rigorous definition of decimal expansions here, it can be shown that every real number r can be expressed in the form $\textstyle r = \sum_{n=0}^{\infty} \frac{a_n}{10^n}$ where an ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
• Consider the series $\textstyle \tfrac{1}{1\cdot 2}+\tfrac{1}{2\cdot 3}+\tfrac{1}{3\cdot 4}\cdots=\sum_{n=1}^\infty\tfrac{1}{n(n+1)}$. At first it may seem difficult to determine the partial sums and decide whether or not the series converges. It turns out this series is nicer then it first appears, this is because $\textstyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$ Thus

$\sum_{n=1}^N \frac{1}{n(n+1)}=\sum_{n=1}^N \frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{N}-\frac{1}{N+1}=1-\frac{1}{N+1}.$

Thus $\textstyle \lim_{N\to\infty}\sum_{n=1}^N\tfrac{1}{n(n+1)}=\lim_{N\to\infty}(1-\tfrac{1}{N+1})=1.$ This is an example of a telescoping series, that is a series that can be written in the form where the next term cancels with the previous terms. That is,
$\sum_{n=1}^{\infty}a_n-a_{n+1}$
For all such series the partial sums are given by SN = a1 − aN+1, that is the terms of the series collapse like a telescope, leaving only the first and last. Such a series converges to a1 − lim aN+1.
• Another important example of a series is a the geometric series given by $\sum_{n=0}^{\infty} r^n$. The partial sums for the geometric series first appear quite complicated, at first glance the partial sums would just appear to be SN = 1 + r +r2 + … + rN, but if we calculate (1 − r)SN then the sum telescopes and we are left with (1 − r)SN = (1 − r)N+1. Notice that this sum holds for any r or N, so we conclude that $\textstyle S_N=\tfrac{1-r^{N+1}}{1-r}$. Thus a geometric series converges if |r| < 1.

## Basic FactsEdit

Here we collect facts about series that follow immediately from the properties of finite sums a limits and do not require a delicate analysis of the limits involved.

### Theorem (Algebraic Operations)Edit

Suppose ∑ an and ∑ bn are convergent series, then we immediately have the following two theorems.

• For any real number c the series ∑ (c·an) is convergent and converges to c·(∑ an).
• The series ∑ (an + bn) is convergent and converges to (∑ an ) + (∑ bn )

#### ProofEdit

For the first statement notice that for any natural number N, we have that $\textstyle \sum_{n=1}^N c\cdot a_n=\textstyle c\cdot \sum_{n=1}^N a_n$. Thus

$\lim_{N\to\infty}\sum_{n=1}^N c\cdot a_n=\lim_{N\to\infty}c\cdot \sum_{n=1}^N a_n=c\cdot \lim_{N\to\infty}\sum_{n=1}^N a_n$.

This proves the first statement.

For the second statement again notice that for any N we have that $\textstyle \sum_{n=1}^N a_n+b_n=\sum_{n=1}^N a_n+\sum_{n=1}^N b_n$. Thus

$\lim_{N\to\infty}\sum_{n=1}^N a_n+b_n=\lim_{N\to\infty}\left(\sum_{n=1}^N a_n+\sum_{n=1}^N b_n\right)=\lim_{N\to\infty}\sum_{n=1}^N a_n+\lim_{N\to\infty}\sum_{n=1}^N b_n$.

This proves the second statement. $\Box$

The following is merely a restatement of Cauchy's Criterion.

### Theorem (Cauchy's Criterion for Series)Edit

A series $\textstyle \sum_{n=1}^{\infty} a_n$ converges if and only if for any ε > 0 there is a natural number N so that $\textstyle |\sum_{k=n+1}^m a_k| < \epsilon$ for any natural numbers m > n > N.

#### ProofEdit

By the definition of convergence for a series $\textstyle \sum_{n=1}^{\infty} a_n$ converges if and only if $\textstyle\lim_{N\to\infty}\sum_{k=1}^N a_k$ converges. But by the Cauchy criterion for sequences this limit exists if and only if for any ε > 0 we can find a natural number N so that for nm > N we have that

$\left|\sum_{k=1}^n a_k-\sum_{k=1}^m a_k\right|<\epsilon.$

Without loss of generality we may assume m > n. Canceling the terms that occurs in both sums completes the proof. $\Box$

## Absolute ConvergenceEdit

As we will see series may behave in ways that defy our intuition. So, it is useful to identify which classes of convergent series behave in a way more consistent with our intuition. We shall see that once such class of series are the so called absolutely convergent series.

Definition A series $\textstyle \sum_{n=1}^{\infty} a_n$ converges absolutely if the series of the absolute values of its terms converges, that is, if $\textstyle\sum_{n=1}^{\infty} |a_n|$ is a convergent series.

Notice we have not said required that ∑ an is convergent, this is because the property of absolute convergence implies that ∑ an is convergent.

### Theorem (Absolute Convergence)Edit

If $\textstyle\sum_{n=1}^{\infty} a_n$ converges absolutely, then it converges.

#### ProofEdit

Suppose $\textstyle\sum_{n=1}^{\infty} a_n$ is absolutely convergent. By the Cauchy criterion we can show that $\textstyle \sum_{n=1}^\infty a_n$ converges if for any ε > 0 we can find an N so that if m > n > N then $\textstyle |\sum_{k=n+1}^{m} a_k|<\epsilon$. But we know that $\textstyle \sum_{n=1}^\infty |a_n|$ converges, so ε > 0 by the Cauchy criterion choose N so that if m > n > N then $\textstyle |\sum_{k=n+1}^{m} |a_k||=\textstyle \sum_{k=n+1}^{m} |a_k|<\epsilon$. With this N in hand take any m > n > N, it then follows from the triangle inequality that

$|\sum_{k=n+1}^m a_k|\leq \sum_{k=n+1}^m |a_k|< \epsilon.$

Therefore by the Cauchy criterion $\sum_{n=1}^\infty a_n$ converges. $\Box$

## Tests for convergence and divergenceEdit

Given a series $\textstyle \sum_{n=1}^\infty a_n$ it is very useful to be able to tell if the series converges or diverges. Particularly if we can do so simply by looking at the terms. In this section we set out to collect such theorems. Notice we have already seen one example. That is a series is convergent if it is absolutely convergent, now we explore when we can decide if a series is convergent.

Although the following theorem is stated in terms of convergence, it actually gives a useful test for divergence. Namely, if the terms of a series do not have limit 0, the series must diverge.

### Theorem (Terms Have Limit 0)Edit

For any converging series $\textstyle\sum_{n=1}^{\infty} a_n$ the terms must tend to 0, that is $\textstyle \lim_{n\rightarrow \infty} a_n = 0.$

#### ProofEdit

Given any ε > 0, by the Cauchy Criterion for series, we can choose a natural number N so that $\textstyle |\sum_{i=n+1}^m a_i|<\epsilon$ when ever m > n ≥ N. In particular, for any k > N we may apply this in the case when n = k − 1 and m = k. In this case the sum reduces to the single term ak. Thus if k > N we have that |ak| < ε and therefore the sequence (an) → 0. $\Box$

Remark To give an example of how to use this to test for divergence of a series consider the example of a geometric series $\textstyle \sum_{n=0}^\infty r^n$. We have already shown that this series converges if |r| < 1. On the other hand we have not determined if the series converges or diverges for other values of r. Now it is clear that if |r| ≥ 1 then the sequence rn does not tend to zero. So we now know that the geometric series $\textstyle \sum_{n=0}^\infty r^n$ converges if and only of |r| < 1.

### Theorem (Positive Series Converge)Edit

Suppose $\textstyle \sum_{n=1}^{\infty} a_n$ is a series of non-negative terms, that is an ≥ 0, then the series either converges if and only if the partial sums are bounded above.

#### ProofEdit

Since the terms are non-negative, we clearly have $\textstyle \sum_{n=1}^N a_n \leq \sum_{n=1}^{N+1} a_n$ and hence the partial sums form a monotonic sequence. If the partial sums are bounded then the series converges. If they are unbounded, then for any M > 0 we can find an N so that $\textstyle \sum_{n=1}^N a_n \geq M$, since the partial sums are non-decreasing, we have that $\textstyle\lim_{N \rightarrow \infty}\sum_{n=1}^N a_n = \infty$, and hence the series does not converge. $\Box$

Remark It follows from the proof, together with the Convergence of monotone sequences, that if a sequence of non-negative terms converges we may take the sum of the sequence as an upper bound for the partial sums, as our intuition would dictate.

To determine whether or not a series converges, sometimes it is useful to compare it term-by-term with another series whose convergence is understood. The following theorem gives one such method for comparing.

### Theorem (Comparison Test)Edit

Suppose that 0 ≤ an ≤ bn for all natural numbers n and consider the series $\textstyle \sum_{n=1}^{\infty} a_n$ and $\textstyle \sum_{n=1}^{\infty} b_n$. If the $\textstyle \sum_{n=1}^{\infty} b_n$ converges, then $\textstyle \sum_{n=1}^{\infty} a_n$ converges. Furthermore, if $\textstyle \sum_{n=1}^{\infty} a_n$ diverges, then $\textstyle \sum_{n=1}^{\infty} b_n$ diverges.

#### ProofEdit

First suppose that the ∑ bn converges. Then by our previous theorem we know that the partial sums for some real number we know that $\textstyle\sum_{n=1}^N b_n\leq\sum_{n=1}^\infty b_n$. Since an ≤ bn if follows that

$\sum_{k=1}^{N} a_n \leq \sum_{k=1}^{N} b_n \leq \sum_{n=1}^{\infty} b_n$.

Thus the partial sums of ∑ an are bounded above, since ∑ an is also a series of non-negative terms it follows that ∑ an converges.

Now suppose that ∑ an diverges. Since it is a series of non-negative terms the pervious theorem tells us that for any real number M we can find an N so that $\textstyle M<\sum_{n=1}^N a_n$. Since an ≤ bn we have that

$M<\sum_{k=1}^{N} a_n \leq \sum_{k=1}^{N} b_n$.

Therefore the partial sums of ∑ bn cannot be bounded above, and hence by the previous theorem ∑ bn diverges. $\Box$

### Theorem (Limit Comparison Test)Edit

Suppose $\textstyle \sum_{n=1}^{\infty} a_n$ and $\textstyle \sum_{n=1}^{\infty} b_n$ series of positive terms so that $\textstyle 0\leq\limsup_{n\to\infty} \tfrac{a_n}{b_n}<\infty$. In this case, if $\textstyle \sum_{n=1}^{\infty} b_n$ converges, then $\textstyle \sum_{n=1}^{\infty} a_n$ converges.

#### ProofEdit

Suppose the limit of the absolute value of their ratios converges to some positive number r < ∞. Then there exists an N such that if n > N, that $|\,|\tfrac{a_n}{b_n}|-r|<\tfrac{r}{2}$, so $\tfrac{r}{2}<|\tfrac{a_n}{b_n}|<\tfrac{3r}{2}$. This means that $\tfrac{r}{2}|b_n|<|a_n|<\tfrac{3r}{2}|b_n|$. Hence, by the comparison test, if $\sum_{n=1}^{\infty} b_n$ converges, then $\sum_{n=1}^{\infty} a_n$ also converges. Dividing by $\tfrac{r}{2}$, one can also see by the comparison test that if $\sum_{n=1}^{\infty} a_n$ converges, then $\sum_{n=1}^{\infty} b_n$ also converges.

We need to have ways of checking for convergence and divergence of various series. The following types of series arise quite often, and it is easy to verify when they converge and diverge:

### Theorem (Ratio Test)Edit

Supoose that $\textstyle \lim_{n \rightarrow \infty} \left|\frac{x_{n+1}}{x_n}\right|$ converges and equals some real number r. If r < 1, then $\textstyle \sum_{n=1}^{\infty} x_n$ converges absolutely. If r > 1, then $\textstyle \sum_{n=1}^{\infty} x_n$ diverges. Finally, if r = 1, then $\textstyle \sum_{n=1}^{\infty} x_n$ may either converge or diverge.

#### ProofEdit

First suppose that r < 1. Let ε = 1 − r, since r < 1 it follows that ε > 0.

Since $\textstyle\lim_{n \rightarrow \infty} \left|\frac{x_{n+1}}{x_n}\right| = r$, we can choose an N so that | |xn+1/xn| − r| < ε/2 for all n ≥ N. In particular, if n ≥ N, then |xn+1|/|xn'| < r + ε/2. Notice that our particular choice of ε guarantees that the ratio r + ε/2 is less than 1.

Thus

$\sum_{n=1}^{\infty} |x_n|$ = $\sum_{n=1}^{N} |x_n| + \sum_{n=N+1}^{\infty} |x_n|$
$= \sum_{n=1}^N |x_n| + \sum_{n=N+1}^{\infty} |x_N|\frac{|x_{N+1}|}{|x_N|}\frac{|x_{N+2}|}{|x_{N+1}|}\cdots\frac{|x_{n}|}{|x_{n-1}|}$
$\leq \sum_{n=1}^N |x_n| + |x_N|\sum_{n=N+1}^{\infty}(r + \frac{\epsilon}{2})^{n-N}.$

The last series converges because it is a geometric series whose ratio, r + ε/2 , is less than 1.

By the comparison test, $\textstyle\sum_{n=1}^{\infty} |x_n|$ converges. Thus $\textstyle\sum_{n=1}^{\infty} x_n$ converges absolutely. $\Box$

The next theorem, the root test, is stronger than the ratio test in the sense that it works whenever the ratio test works (and returns the same number r), and it sometimes works even when the ratio test does not.

### Theorem (Root Test)Edit

Let $R = \limsup_{n\rightarrow\infty}(|a_n|^{\frac{1}{n}})$. If R < 1, then the series $\sum_{n=1}^{\infty}a_n$ converges absolutely. If R > 1, then it diverges.

#### ProofEdit

If R<1, let $R<\rho<1$. Since $\limsup_{n\rightarrow\infty}(|a_n|^{\frac{1}{n}}) = R$ $\exists N: \forall n \geq N: |\sup\{|a_k|^{\frac{1}{k}}|k > n\} - R| < \rho - R$.

That is, $\forall n \geq N: \sup\{|a_k|^{\frac{1}{k}}|k > n\} < \rho$

Thus $\sum_{n=1}^{\infty}|a_n|$$= \sum_{n=1}^{\infty}(|a_n|^{\frac{1}{n}})^n$$= \sum_{n=1}^{N-1}|a_n| + \sum_{n=N}^{\infty}(|a_n|^{\frac{1}{n}})^n$$< \sum_{n=1}^{N-1}|a_n| + \sum_{n=N}^{\infty}(\sup\{|a_k|^{\frac{1}{k}}|k > n \})^n$$< \sum_{n=1}^{N-1}|a_n| + \sum_{n=N}^{\infty}(\rho)^n$, which converges since it is a constant plus a geometric series.

By the comparison test, $\sum_{n=1}^{\infty}|a_n|$ converges as well. Thus $\sum_{n=1}^{\infty}a_n$ converges absolutely.

If R > 1, then $\sup\{|a_n|^{\frac{1}{n}} | k>n\} > 1$. Thus there are infinitely many n such that $|a_n|^{\frac{1}{n}} > 1 \implies |a_n| > 1$.

Thus $(a_n) \not\rightarrow 0$, which implies that $\sum_{n=1}^{\infty}a_n$ diverges. $\Box$

We'll often be asked to consider series of the form $\sum_{n=1}^{\infty} a_nb_n$. The following theorems give criteria for convergence of these series.

### Theorem (Dirichlet's Test)Edit

If the partial sums $\sum_{n=1}^{N}x_n$ are bounded, and $(y_n)$ is a decreasing sequence with $\lim_{n\rightarrow\infty} y_n = 0$, then $\sum_{n=1}^{\infty}x_ny_n$ converges. [Note: $\sum_{n=1}^{\infty}x_n$ need not converge]

#### ProofEdit

Let $s_n = \sum_{m=1}^n x_m$ be the $n$th partial sum, so that there exists $B>0$ such that $|s_n| \le B$.

We can write

$\sum_{n=1}^N x_ny_n = x_1y_1 + \sum_{n=2}^N (s_n-s_{n-1})y_n = x_1y_1 + \sum_{n=2}^N s_ny_n - \sum_{n=2}^N s_{n-1}y_n$.

Changing the index of summation in the last sum, this becomes

$\sum_{n=1}^N x_ny_n = x_1y_1 + \sum_{n=2}^N s_ny_n - \sum_{n=1}^{N-1} s_ny_{n+1} = x_1y_1 - x_1y_2 + \sum_{n=2}^{N-1} s_n(y_n-y_{n+1}) + s_Ny_N$.

The sum on the right-hand side is bounded absolutely by the telescoping sum

$\sum_{n=2}^{N-1} |s_n(y_n-y_{n+1})| \le B \sum_{n=2}^{N-1} (y_n-y_{n+1}) = B(y_2-y_N) \le By_2$;

here we have used the fact that $(y_n)$ is positive and decreasing. It follows that

$\sum_{n=2}^\infty s_n(y_n-y_{n+1})$ is absolutely convergent, hence convergent.

Notice that $\lim_{N\to\infty} s_Ny_N = 0$ since $s_N$ is bounded and $(y_n)$ tends to 0 as $n$ tends to infinity. Therefore we can take limits as $N$ goes to infinity:

$\sum_{n=1}^\infty x_ny_n = x_1y_1 - x_1y_2 + \sum_{n=2}^\infty s_n(y_n-y_{n+1}) + \lim_{N\to\infty} s_Ny_N = x_1y_1 - x_1y_2 + \sum_{n=2}^\infty s_n(y_n-y_{n+1})$,

which proves that the left-hand side is convergent.

Abel's Test can be regarded as a special case of Dirichlet's Test, once some modifications have been made:

### Theorem (Abel's Test)Edit

If $\sum_{n=1}^{\infty}x_n$ converges and $(y_n)$ is a positive, decreasing sequence, then $\sum_{n=1}^{\infty}x_ny_n$ converges.

#### ProofEdit

Because $(y_n)$ is a bounded, monotone sequence, it converges to some limit y.

Let $z_n = y_n - y$. Then $\sum_{n=1}^{\infty}x_n$ and $(z_n)$ satisfy the conditions for Dirichlet's test.

$\sum_{n=1}^{k}x_ny_n$$= y\sum_{n=1}^{k}x_n + \sum_{n=1}^{k}x_nz_n$. By Dirichlet's test, $\sum_{n=1}^{k}x_nz_n$ converges as $k \rightarrow \infty$.

Since both sums on the right converge, $\sum_{n=1}^{\infty}x_ny_n$ converges as well. $\Box$

## Examples with ProofEdit

Now we'll do the computations promised in the Examples, plus a few extra.

### Theorem (Geometric Series)Edit

If $|r|<1$, the geometric series $\sum_{n=1}^{\infty} a r^n = \frac{a}{1-r}$. If $|r| \geq 1$, the series diverges.

#### ProofEdit

In this case, it is best to explicitly compute the partial sums and take the limit. Without loss of generality, we'll consider the case $a = 1$. Then we can apply the theorem on algebraic operations to obtain the general result.

Note that $s_n(1-r) = (1 + r + r^2 + ... + r^n)(1-r) = (1 + r + r^2 + ... + r^n) - (r + r^2 + ... + r^{n+1}) = 1 - r^{n+1}$

Thus $s_n = \frac{1-r^n}{1-r}$. Taking the limit (and remembering some basic facts about sequences):

If $|r| < 1$, $\sum_{n=1}^{\infty} r^n = \lim_{n \rightarrow \infty} s_n = \lim_{n \rightarrow \infty} \frac{1-r^n}{1-r} = \frac{1}{1-r}$.

If $|r| \geq 1$, the sequence of partial sums diverges to infinity, and thus by definition the series diverges.

This proof seems to work except for the fact that this series will converge to $\frac{a} {1-r}$ $\Box$

### Theorem (p-series)Edit

The p-series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges for $p>1$ and diverges for $0. [Note: We have technically only defined this series for p rational. However, the theorem is still valid when p is irrational, for the same reasons]

#### ProofEdit

First we'll consider the special cases $p = 1, p = 2$, and then obtain the general result from these.

• If p = 1, let $x_n$ be the greatest power of 2 less than $\frac{1}{n}$. That is, $(x_n) = (\frac{1}{2}, \frac{1}{4},\frac{1}{4}, \frac{1}{8}, \frac{1}{8},\frac{1}{8},\frac{1}{8},\dots)$.

Grouping like terms together, $\sum_{i=1}^{\infty} x_n$$= \sum_{i=0}^{\infty}(\sum_{j=2^i}^{2^{i+1}-1}x_j)$$= \sum_{i=0}^{\infty}(\sum_{j=2^i}^{2^{i+1}-1} \frac{1}{2^{i+1}})$$= \sum_{i=1}^{\infty} \frac{1}{2}$, which diverges.

By definition, $x_n < \frac{1}{n}$, so by the comparison test $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges.

• If p = 2, let $x_n = \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$ if n>1 and 1 if n=1.

Using the theorem on telescoping series, $\sum_{n=1}^{\infty} x_n$$= 1 + \sum_{n=2}^{\infty}(\frac{1}{n-1} - \frac{1}{n})$$= 1 + 1 - \lim_{n\rightarrow\infty}(1/n) = 2$

By definition, $x_n > \frac{1}{n \cdot n} = \frac{1}{n^2}$, so by the comparison test $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges as well.

• If 0 < p < 1, then $\frac{1}{n^p} > \frac{1}{n}$. By the comparison test $\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges.

### Theorem (Decimal Expansions)Edit

Given any real number x, $0 there is a unique sequence $(x_n): \sum_{n=0}^{\infty} \frac{x_n}{10^n} = x$, $0 \leq x_n<10$ and $\forall N: \exists n \geq N: x_n \not= 9$

#### ProofEdit

Inductively, assume that there exist $x_0, x_1, \dots, x_k$ such that $\sum_{n=0}^{k} \frac{x_n}{10^n} \leq x < \sum_{n=0}^{k} \frac{x_n}{10^n} + \frac{1}{10^k}$.

Rearranging, we see that $0 \leq x - \sum_{n=0}^{k} \frac{x_n}{10^n} < \frac{1}{10^k}$, or $0 \leq (x - \sum_{n=0}^{k} \frac{x_n}{10^n})10^{k+1} < 10$.

Let $x_{k+1}$ by the greatest integer such that $x_{k+1} \leq (x - \sum_{n=0}^{k} \frac{x_n}{10^n})10^{k+1}$.

Then $\frac{x_{k+1}}{10^{k+1}} \leq x - \sum_{n=0}^{k} \frac{x_n}{10^n} < \frac{x_{k+1}+1}{10^{k+1}}$(otherwise, $x_{k+1}$ would not be the greatest).

Adding $\sum_{n=0}^{k} \frac{x_n}{10^n}$, we see that $\sum_{n=0}^{k+1} \frac{x_n}{10^n} \leq x < \sum_{n=0}^{k+1} \frac{x_n}{10^n} + \frac{1}{10^{k+1}}$.

Given $\epsilon>0$ pick N such that $\epsilon > \frac{1}{10^N}$.

Thus $|\sum_{n=0}^{k} \frac{x_n}{10^n} -1| < \frac{1}{10^k} < \epsilon$ for all $k > N$. That is, $x = \sum_{n=0}^{\infty} \frac{x_n}{10^n}$.

TODO: eliminate 9s, uniqueness.

Topics to come: Rearrangement of terms, Alternating Series Test, Sums of products(i.e. Abel's Test and Dirichlet's Test), Multiple Summations, Infinite Products, decimal expansions, zeta function

## Rearrangement of termsEdit

### DefinitionEdit

By a rearrangement of terms, we mean a bijection f from the natural numbers to itself. Then a rearrangement of a series $\sum_{i=0}^\infty a_n$ is any series $\sum_{i=0}^\infty a_f(n)$ where f is any rearrangement.

### TheoremEdit

If a series is absolutely convergent, then all rearrangements of terms converge to the same limit.

### TheoremEdit

If a series $\{a_n\}$ is conditionally convergent but not absolutely convergent, then there exists a rearrangement for any a<b such that lim inf $\sum \limits_{i=0}^n a_n=a$ and such that lim sup $\sum\limits_{i=0}^n a_n=b$