Putting it all together

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The various concepts associated with sequences and series are reviewed.

All of the series convergence tests we have used require that the underlying sequence $\seq {a_n}$ be a positive sequence. We can actually relax this and state that there must be an $N>0$ such that $a_n>0$ for all $n>N$ ; that is, $\seq {a_n}$ is positive for all but a finite number of values of $n$ . We’ve also stated this by saying that the tail of the sequence must have positive terms. In this section we explore series whose summation includes negative terms.

Alternating series test

We start with a very specific form of series, where the terms of the summation alternate between being positive and negative.

Let $\seq {a_n}$ be a positive sequence. An alternating series is a series of either the form $\sum _{n=1}^\infty (-1)^na_n\qquad \text {or}\qquad \sum _{n=1}^\infty (-1)^{n+1}a_n.$

In essence, the signs of the terms of $\seq {a_n}$ alternate between positive and negative.

Recall that the terms of the harmonic series come from the harmonic sequence $\seq {a_n} = \seq {1/n}$ . An important alternating series is the alternating harmonic series: $\sum _{n=1}^\infty \frac {(-1)^{n+1}}{n} = 1-\frac 12+\frac 13-\frac 14+\frac 15-\frac 16+\cdots$

Geometric series are also alternating series when $r<0$ . For instance, if $r=-1/2$ , the geometric series is $\sum _{n=0}^\infty \left (\frac {-1}{2}\right )^n = 1-\frac 12+\frac 14-\frac 18+\frac 1{16}-\frac 1{32}+\cdots$

We know that geometric series converge when $|r|<1$ and have the sum: $\sum _{n=0}^\infty r^n = \frac 1{1-r}.$ When $r=-1/2$ as above, we find $\sum _{n=0}^\infty \left (\frac {-1}{2}\right )^n = \frac 1{1-(-1/2)} = \frac 1{3/2} = \frac 23.$

A powerful convergence theorem exists for other alternating series that meet a few conditions.

Alternating Series Test Let $\seq {a_n}$ be a positive, nonincreasing sequence where $\lim _{n\to \infty }a_n=0$ . Then $\sum _{n=1}^\infty (-1)^{n}a_n \qquad \text {and}\qquad \sum _{n=1}^\infty (-1)^{n+1}a_n$ converge.

Does the alternating series test apply to the series $\sum _{n=1}^\infty \frac {(-1)^{n+1}}{n} ?$

yes no

This is the alternating harmonic series as seen previously. The underlying sequence is $\seq {a_n} = \seq {1/n}$ , which is positive, decreasing, and approaches 0 as $n\to \infty$ . Therefore we can apply the alternating series test and conclude this series converges.

Does the alternating series test apply to the series $\sum _{n=1}^\infty \frac {(-1)^n\ln n}{n}?$

yes no

The underlying sequence is $\seq {a_n} = \seq {\ln n/n}$ . This is positive and approaches $0$ as $n\to \infty$ (use L’Hôpital’s Rule). However, the sequence is not decreasing for all $n$ . It is straightforward to compute $a_1=0$ , $a_2\approx 0.347$ , $a_3\approx 0.366$ , and $a_4\approx 0.347$ : the sequence is increasing for at least the first $3$ terms.

However, we do not “give-up” and immediately conclude that we cannot apply the alternating series test. Rather, consider the long term behavior of $\seq {a_n}$ . Treating $a_n=a(n)$ as a continuous function of $n$ defined on $(1,\infty )$ , we can take its derivative: $a'(n) = \frac {1-\ln n}{n^2}.$ The derivative is negative for all $n\geq 3$ (actually, for all $n>e$ ), meaning $a(n)=a_n$ is decreasing on $(3,\infty )$ . We can now apply the alternating series test to the series when we start with $n=3$ and conclude that $\sum _{n=3}^\infty (-1)^n\frac {\ln n}{n}$ converges; adding the terms with $n=1$ and $n=2$ do not change the convergence.

The important lesson here is that as before, if a series fails to meet the criteria of the alternating series test on only a finite number of terms, we can still apply the test.

Does the alternating series test apply to the series $\sum _{n=1}^\infty (-1)^{n+1}\frac {|\sin n|}{n^2}?$

yes no

The underlying sequence is $\seq {a_n} = |\sin n|/n$ . This sequence is positive and approaches $0$ as $n\to \infty$ . However, it is not a decreasing sequence; the value of $|\sin n|$ oscillates between $0$ and $1$ as $n\to \infty$ . We cannot remove a finite number of terms to make $\seq {a_n}$ decreasing, therefore we cannot apply the alternating series test.

Keep in mind that this does not mean we conclude the series diverges; in fact, it does converge. We are just unable to conclude this based on the alternating series test.

Approximating alternating series

While there are many factors involved when studying rates of convergence, the alternating structure of an alternating series gives us a powerful tool when approximating the sum of a convergent series.

Alternating Series Approximation Let $\seq {a_n}$ be a sequence that satisfies the hypotheses of the alternating series test, let $S_n$ be the $n$ th partial sum, and let $L = \sum _{n=1}^\infty (-1)^{n}a_n\qquad \text {or}\qquad L=\sum _{n=1}^\infty (-1)^{n+1}a_n.$ Then

$|L-S_n| \leq a_{n+1}$ , and
$L$ is between $S_n$ and $S_{n+1}$ .

In this case, $R_n=L-S_n$ is called the $n$ th remainder of the series.

Here is the basic idea behind this theorem. Say we have an alternating sequence, $\seq {a_n}$ . Let’s assume the first term is positive, so the second is negative, and so on. We add the first two numbers and get some number $S_2$ . Now $S_2$ is smaller than $S_1 = a_1$ , because we subtracted something from $a_1$ . Next, we add on the third term, $a_3$ , to get the partial sum $S_3$ . This $S_3$ is bigger than $S_2$ , because the sequence is alternating, but is smaller than $S_1$ , because the sequence is decreasing.

If we know the series converges to some $L$ , we can see that we must be bouncing back and forth around $L$ as we add and subtract terms. At one point, $S_n$ is larger than $L$ , and then subtracting off the next term makes the partial sum smaller than $L$ . In other words, the true limit $L$ must be between $S_n$ and $S_{n+1}$ . Imagine plotting $\seq {a_n}$ , $\seq {S_n}$ , and $L$ on a number line. (Or, try it yourself with the alternating harmonic series!) can be no further from $S_n$ than whatever the next term in the sequence is. How do we get from $S_n$ to $S_{n+1}$ ? By adding (or subtracting) $a_{n+1}$ , which takes us back “across” $L$ again. In other words, the distance between $L$ and $S_n$ can be no more than $a_{n+1}$ .

See if you can use these same ideas to prove the alternating series test!

Let’s see an example of approximating an alternating series.

Approximate the sum of the alternating harmonic series with an error less than $10^{-2}$ .

Look at the $(n+1)$ th term. $\begin{align*} \frac{1}{n+1} &< 10^{-2}\\ n+1&> 100\\ n&> 99 \end{align*}$

Using a computer, we can see that $S_{99} = 0.69849462\dots$ We will see later that the true value of this series is $\ln (2)$ . Comparing

$\begin{align*} \ln(2) &= 0.69314718\dots \qquad \text{and}\\ S_{99} &= 0.69849462\dots \end{align*}$

we have our desired accuracy.

Absolute convergence versus conditional convergence

It is an interesting result that the harmonic series, $\sum _{n=1}^\infty \frac 1n$ diverges, yet the alternating harmonic series, $\sum _{n=1}^\infty (-1)^{n+1}\frac 1n,$ converges. The notion that simply alternating the signs of the terms in a series can change a series from divergent to convergent leads us to the following definitions.

A series $\sum _{n=1}^\infty a_n$ converges absolutely if $\sum _{n=1}^\infty |a_n|$ converges.
A series $\sum _{n=1}^\infty a_n$ converges conditionally if $\sum _{n=1}^\infty a_n$ converges but $\sum _{n=1}^\infty |a_n|$ diverges.

Note, in the definition above, $\sum _{n=1}^\infty a_n$ is not necessarily an alternating series; it may just have some negative terms.

Does the series $\sum _{n=1}^\infty (-1)^n\frac {n+3}{n^2+2n+5}$ converge absolutely, converge conditionally, or diverge?

The series converges conditionally. The series converges absolutely. The series diverges.

We can show the series $\sum _{n=1}^\infty \left |(-1)^n\frac {n+3}{n^2+2n+5}\right |= \sum _{n=1}^\infty \frac {n+3}{n^2+2n+5}$ diverges using the limit comparison test, comparing with $1/n$ . The series $\sum _{n=1}^\infty (-1)^n\frac {n+3}{n^2+2n+5}$ converges using the alternating series test; we conclude the series converges conditionally.

Does the series $\sum _{n=1}^\infty (-1)^n\frac {n^2+2n+5}{2^n}$ converge absolutely, converge conditionally, or diverge?

The series converges conditionally. The series converges absolutely. The series diverges.

We can show the series $\sum _{n=1}^\infty \left |(-1)^n\frac {n^2+2n+5}{2^n}\right |=\sum _{n=1}^\infty \frac {n^2+2n+5}{2^n}$ converges using the ratio test. Therefore we conclude $\sum _{n=1}^\infty (-1)^n\frac {n^2+2n+5}{2^n}$ converges absolutely.

Does the series $\sum _{n=3}^\infty (-1)^n\frac {3n-3}{5n-10}$ converge absolutely, converge conditionally, or diverge?

The series converges conditionally. The series converges absolutely. The series diverges.

The series $\sum _{n=3}^\infty \left |(-1)^n\frac {3n-3}{5n-10}\right | = \sum _{n=3}^\infty \frac {3n-3}{5n-10}$ diverges using the divergence test, so it does not converge absolutely. The series $\sum _{n=3}^\infty (-1)^n\frac {3n-3}{5n-10}$ fails the conditions of the alternating series test as $(3n-3)/(5n-10)$ does not approach $0$ as $n\to \infty$ . We can state further that this series diverges; as $n\to \infty$ , the series effectively adds and subtracts $3/5$ over and over. This causes the sequence of partial sums to oscillate and not converge. Therefore the series diverges.

Knowing that a series converges absolutely allows us to make two important statements. The first, given in the following theorem, is that absolute convergence is “stronger” than regular convergence. That is, just because $\sum _{n=1}^\infty a_n$ converges, we cannot conclude that $\sum _{n=1}^\infty |a_n|$ will converge, but knowing a series converges absolutely tells us that $\sum _{n=1}^\infty a_n$ will converge.

One reason this is important is that our convergence tests all require that the underlying sequence of terms be positive. By taking the absolute value of the terms of a series where not all terms are positive, we are often able to apply an appropriate test and determine absolute convergence. This, in turn, determines that the series we are given also converges.

The second statement relates to rearrangements of series. When dealing with a finite set of numbers, the sum of the numbers does not depend on the order in which they are added. (So $1+2+3 = 3+1+2$ .) One may be surprised to find out that when dealing with an infinite set of numbers, the same statement does not always hold true: some infinite lists of numbers may be rearranged in different orders to achieve different sums. The theorem states that the terms of an absolutely convergent series can be rearranged in any way without affecting the sum.

Absolute Convergence Let $\sum _{n=1}^\infty a_n$ be a series that converges absolutely. Let $\seq {b_n}$ be any rearrangement of the sequence $\seq {a_n}$ . Then $\sum _{n=1}^\infty b_n = \sum _{n=1}^\infty a_n.$

This theorem states that rearranging the terms of an absolutely convergent series does not affect its sum. Making such a statement implies that perhaps the sum of a conditionally convergent series can change based on the arrangement of terms. Indeed, it can. The Riemann rearrangement theorem (named after Bernhard Riemann) states that any conditionally convergent series can have its terms rearranged so that the sum is any desired value, including $\infty$ !

As an example, consider the alternating harmonic series once more. We have stated that $\sum _{n=1}^\infty \frac {(-1)^{n+1}}{n} =1-\frac 12+\frac 13-\frac 14+\frac 15-\frac 16+\frac 17\cdots = \ln 2.$ Consider the rearrangement where every positive term is followed by two negative terms: $1-\frac 12-\frac 14+\frac 13-\frac 16-\frac 18+\frac 15-\frac 1{10}-\frac 1{12}\cdots$ (Convince yourself that these are exactly the same numbers as appear in the alternating harmonic series, just in a different order.) Now group some terms and simplify:

$\begin{align*} \left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\frac1{12}+\cdots &= \\ \frac12-\frac14+\frac16-\frac18+\frac1{10}-\frac{1}{12}+\cdots &= \\ \frac12\left(1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots\right) & = \frac12\ln 2. \end{align*}$

By rearranging the terms of the series, we have arrived at a different sum!

One could try to argue that the alternating harmonic series does not actually converge to $\ln 2$ , and here is an example of such an argument. According to the alternating series test, we know that this series converges to some number $L$ . If, as our intuition tells us should be true, the rearrangement does not change the sum, then we have just seen that $L = L/2$ . The only possibility for $L$ is then $L=0$ . But the alternating series approximation theorem quickly shows that $L>0$ . The only conclusion is that the rearrangement did, contrary to our intuition, change the sum.

The fact that conditionally convergent series can be rearranged to equal any number is really an incredible result.

While series are worthy of study in and of themselves, our ultimate goal within calculus is the study of power series, which we will consider in the next section. We will use power series to create functions where the output is the result of an infinite summation.