Series

Frequently in analysis it is useful to consider the sum of infinitely many numbers. So for some sequence of numbers (a_n) 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.

Definition

We begin with a sequence (a_n) 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 a_n 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 $\textstyle \lim _{n\to \infty }a_{n}$ does not exist - 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 a_n does not converge. However, the meaning of the previous statement is perfectly clear.

Definition The n-th partial sum of a sequence a_n is defined to be the sum of the first n terms of (a_n), that is

$S_{n}=\sum _{i=1}^{n}a_{i}=a_{1}+a_{2}+\cdots +a_{n}$.

When the sequence a_n is being thought of as the terms of a series, then S_n 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 S_n 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 a_n 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 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 ∑a_n for $\textstyle \sum _{n=1}^{\infty }a_{n}$

Examples

• The notion of an infinite sum can be a little more subtle than it first appears. For example, consider the sequence a_n = (−1)ⁿ. Does the sum of the a_n converge or diverge? We could consider the partial sums S_N, 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 $\textstyle r=\sum _{n=0}^{\infty }{\frac {a_{n}}{10^{n}}}$ where a_n ∈ {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. However, it turns out that this series is nicer than it first appears, since $\textstyle {\frac {1}{n(n+1)}}={\frac {1}{n}}-{\frac {1}{n+1}}.$ 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 such that 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 S_N = a₁ − a_N+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 a_N+1.
• Another important example of a series is 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 S_N = 1 + r +r² + … + r^N, but if we calculate (1 − r)S_N then the sum telescopes and we are left with (1 − r)S_N = (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 Facts

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)

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

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

Proof

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. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Box}

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

Theorem (Cauchy's Criterion for Series)

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.

Proof

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 n, m > 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 Convergence

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 ∑ a_n is convergent, this is because the property of absolute convergence implies that ∑ a_n is convergent.

Theorem (Absolute Convergence)

If $\textstyle \sum _{n=1}^{\infty }a_{n}$ converges absolutely, then it converges.

Proof

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 divergence

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)

For any converging series $\textstyle \sum _{n=1}^{\infty }a_{n}$ the terms must tend to 0, that is $\textstyle \lim _{n\rightarrow \infty }a_{n}=0.$

Proof

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 a_k. Thus if k > N we have that |a_k| < ε and therefore the sequence (a_n) → 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 r_n 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 (Nth-term Test)

In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series:

• If $\lim _{n\to \infty }a_{n}\neq 0$ or if the limit does not exist, then $\sum _{n=1}^{\infty }a_{n}$ diverges.

When testing if a series converges or diverges, this test is often checked first due to its ease of use.

Unlike stronger convergence tests, the term test cannot prove by itself that a series converges. In particular, the converse to the test is not true; instead all one can say is:

• If $\lim _{n\to \infty }a_{n}=0,$ then $\sum _{n=1}^{\infty }a_{n}$ may or may not converge. In other words, if $\lim _{n\to \infty }a_{n}=0,$ the test is inconclusive.

Proofs

The test is typically proven in contrapositive form:

If $\sum _{n=1}^{\infty }a_{n}$ converges, then $\lim _{n\to \infty }a_{n}=0.$

• Limit manipulation

If s_n are the partial sums of the series, then the assumption that the series converges means that

$\lim _{n\to \infty }s_{n}=L$

for some number s. Then

$\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }(s_{n}-s_{n-1})=\lim _{n\to \infty }s_{n}-\lim _{n\to \infty }s_{n-1}=L-L=0.$

• Cauchy's criterion

The assumption that the series converges means that it passes Cauchy's convergence test: for every $\varepsilon >0$ there is a number N such that

$\left|a_{n+1}+a_{n+2}+\cdots +a_{n+p}\right|<\varepsilon$

holds for all n > N and p ≥ 1. Setting p = 1 recovers the definition of the statement

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

Theorem (Positive Series Converge)

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

Proof

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)

Suppose that 0 ≤ a_n ≤ b_n 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.

Proof

First suppose that the ∑ b_n 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 a_n ≤ b_n 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 ∑ a_n are bounded above, since ∑ a_n is also a series of non-negative terms it follows that ∑ a_n converges.

Now suppose that ∑ a_n 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 a_n ≤ b_n we have that

$M<\sum _{k=1}^{N}a_{n}\leq \sum _{k=1}^{N}b_{n}$.

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

Theorem (Limit Comparison Test)

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.

Proof

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)

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.

Proof

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 | |x_n+1/x_n| − r| < ε/2 for all n ≥ N. In particular, if n ≥ N, then |x_n+1|/|x_n'| < 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)

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.

Proof

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 $\sum _{n=1}^{\infty }a_{n}b_{n}$. The following theorems give criteria for convergence of these series.

Theorem (Dirichlet's Test)

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 $\sum _{n=1}^{\infty }x_{n}y_{n}$ converges. [Note: $\sum _{n=1}^{\infty }x_{n}$ need not converge]

Proof

Let $s_{n}=\sum _{m=1}^{n}x_{m}$ be the $n$th partial sum, so that there exists $B>0$ such that $|s_{n}|\leq B$.

We can write

$\sum _{n=1}^{N}x_{n}y_{n}=x_{1}y_{1}+\sum _{n=2}^{N}(s_{n}-s_{n-1})y_{n}=x_{1}y_{1}+\sum _{n=2}^{N}s_{n}y_{n}-\sum _{n=2}^{N}s_{n-1}y_{n}$.

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

$\sum _{n=1}^{N}x_{n}y_{n}=x_{1}y_{1}+\sum _{n=2}^{N}s_{n}y_{n}-\sum _{n=1}^{N-1}s_{n}y_{n+1}=x_{1}y_{1}-x_{1}y_{2}+\sum _{n=2}^{N-1}s_{n}(y_{n}-y_{n+1})+s_{N}y_{N}$.

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

$\sum _{n=2}^{N-1}|s_{n}(y_{n}-y_{n+1})|\leq B\sum _{n=2}^{N-1}(y_{n}-y_{n+1})=B(y_{2}-y_{N})\leq By_{2}$;

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 $\lim _{N\to \infty }s_{N}y_{N}=0$ 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:

$\sum _{n=1}^{\infty }x_{n}y_{n}=x_{1}y_{1}-x_{1}y_{2}+\sum _{n=2}^{\infty }s_{n}(y_{n}-y_{n+1})+\lim _{N\to \infty }s_{N}y_{N}=x_{1}y_{1}-x_{1}y_{2}+\sum _{n=2}^{\infty }s_{n}(y_{n}-y_{n+1})$,

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)

If $\sum _{n=1}^{\infty }x_{n}$ converges and $(y_{n})$ is a positive, decreasing sequence, then $\sum _{n=1}^{\infty }x_{n}y_{n}$ converges.

Proof

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_{n}y_{n}$$=y\sum _{n=1}^{k}x_{n}+\sum _{n=1}^{k}x_{n}z_{n}$. By Dirichlet's test, $\sum _{n=1}^{k}x_{n}z_{n}$ converges as $k\rightarrow \infty$.

Since both sums on the right converge, $\sum _{n=1}^{\infty }x_{n}y_{n}$ converges as well. $\Box$

Examples with Proof

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

Theorem (Geometric Series)

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

Proof

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)

The p-series $\sum _{n=1}^{\infty }{\frac {1}{n^{p}}}$ converges for $p>1$ and diverges for $0<p\leq 1$. [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]

Proof

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)

Given any real number x, $0<x<1$ there is a unique sequence $(x_{n}):\sum _{n=0}^{\infty }{\frac {x_{n}}{10^{n}}}=x$, $0\leq x_{n}<10$ and $\forall N:\exists n\geq N:x_{n}\not =9$

Proof

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, $x=\sum _{n=0}^{\infty }{\frac {x_{n}}{10^{n}}}$.

Resources

• Series, Wikibooks: Real Analysis

Licensing

Content obtained and/or adapted from:

• Series, Wikibooks: Real Analysis under a CC BY-SA license