Real Analysis/Series

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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

.

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.

Contents

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
 
where each term an is a real number.

This definition deserves a few comments. The first is 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   does not exist - we have only defined the meaning of the symbols   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
 .
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   instead of Sn when we wish to avoid confusion.

Definition For a series   we define the sum of the series to be the limit of the partial sums. That is, we define:
 .
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   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  . Again this is similar to writing   before the convergence of the sequence an is established. In practice the meaning of the symbols   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 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  

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 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 then the series diverge or converge to 0, according to our theory? 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 must justify why we are allowed to group the +1's together with the -1's without changing the sum. Stated formally, associativity does not necessarily hold for infinite series. 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 of 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   where an ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
  • Consider the series  . At first it may seem difficult to determine the partial sums and decide whether or not the series converges. However, it turns out that this series is nicer than it first appears, since   Thus

     

    Thus   This is an example of a telescoping series - that is, a series that can be written in the form such that the next term cancels with the previous terms. That is,
     
    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 the geometric series given by  . 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  . 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  . Thus

 .

This proves the first statement.

For the second statement again notice that for any N we have that  . Thus

 .

This proves the second statement.  


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

Theorem (Cauchy's Criterion for Series)Edit

A series   converges if and only if for any ε > 0 there is a natural number N so that   for any natural numbers m > n > N.

ProofEdit

By the definition of convergence for a series   converges if and only if   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

 

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

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   converges absolutely if the series of the absolute values of its terms converges, that is, if   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   converges absolutely, then it converges.

ProofEdit

Suppose   is absolutely convergent. By the Cauchy criterion we can show that   converges if for any ε > 0 we can find an N so that if m > n > N then  . But we know that   converges, so ε > 0 by the Cauchy criterion choose N so that if m > n > N then  . With this N in hand take any m > n > N, it then follows from the triangle inequality that

 

Therefore by the Cauchy criterion   converges.  

Tests for convergence and divergenceEdit

Given a series   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   the terms must tend to 0, that is  

ProofEdit

Given any ε > 0, by the Cauchy Criterion for series, we can choose a natural number N so that   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.  

Remark To give an example of how to use this to test for divergence of a series consider the example of a geometric series  . 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   converges if and only of |r| < 1.

Theorem (Positive Series Converge)Edit

Suppose   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   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  , since the partial sums are non-decreasing, we have that  , and hence the series does not converge.  

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   and  . If the   converges, then   converges. Furthermore, if   diverges, then   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  . Since an ≤ bn if follows that

 .

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  . Since an ≤ bn we have that

 .

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

Theorem (Limit Comparison Test)Edit

Suppose   and   series of positive terms so that  . In this case, if   converges, then   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  , so  . This means that  . Hence, by the comparison test, if   converges, then   also converges. Dividing by  , one can also see by the comparison test that if   converges, then   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   converges and equals some real number r. If r < 1, then   converges absolutely. If r > 1, then   diverges. Finally, if r = 1, then   may either converge or diverge.

ProofEdit

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

Since  , 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

  =  
 
 

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

By the comparison test,   converges. Thus   converges absolutely.  


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  . If R < 1, then the series   converges absolutely. If R > 1, then it diverges.

ProofEdit

If R<1, let  . Since    .

That is,  

Thus      , which converges since it is a constant plus a geometric series.

By the comparison test,   converges as well. Thus   converges absolutely.


If R > 1, then  . Thus there are infinitely many n such that  .

Thus  , which implies that   diverges.  


We'll often be asked to consider series of the form  . The following theorems give criteria for convergence of these series.

Theorem (Dirichlet's Test)Edit

If the partial sums   are bounded, and   is a decreasing sequence with  , then   converges. [Note:   need not converge]

ProofEdit

Let   be the  th partial sum, so that there exists   such that  .

We can write

 .

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

 .

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

 ;

here we have used the fact that   is positive and decreasing. It follows that

  is absolutely convergent, hence convergent.

Notice that   since   is bounded and   tends to 0 as   tends to infinity. Therefore we can take limits as   goes to infinity:

 ,

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   converges and   is a positive, decreasing sequence, then   converges.

ProofEdit

Because   is a bounded, monotone sequence, it converges to some limit y.

Let  . Then   and   satisfy the conditions for Dirichlet's test.

  . By Dirichlet's test,   converges as  .

Since both sums on the right converge,   converges as well.  

Examples with ProofEdit

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

Theorem (Geometric Series)Edit

If  , the geometric series  . If  , 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  . Then we can apply the theorem on algebraic operations to obtain the general result.

Note that  

Thus  . Taking the limit (and remembering some basic facts about sequences):

If  ,  .

If  , 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    

Theorem (p-series)Edit

The p-series   converges for   and diverges for  . [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  , and then obtain the general result from these.

  • If p = 1, let   be the greatest power of 2 less than  . That is,  .

Grouping like terms together,     , which diverges.

By definition,  , so by the comparison test   diverges.


  • If p = 2, let   if n>1 and 1 if n=1.

Using the theorem on telescoping series,    

By definition,  , so by the comparison test   converges as well.


  • If 0 < p < 1, then  . By the comparison test   diverges.

Theorem (Decimal Expansions)Edit

Given any real number x,   there is a unique sequence  ,   and  

ProofEdit

Inductively, assume that there exist   such that  .

Rearranging, we see that  , or  .

Let   by the greatest integer such that  .

Then  (otherwise,   would not be the greatest).

Adding  , we see that  .

Given   pick N such that  .

Thus   for all  . That is,  .

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   is any series   where f is any rearrangement.

TheoremEdit

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

ProofEdit

TheoremEdit

If a series   is conditionally convergent but not absolutely convergent, then there exists a rearrangement for any a<b such that lim inf   and such that lim sup  

ProofEdit