
# 11.6: Absolute Convergence and the Ratio and Root Test

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Roughly speaking there are two ways for a series to converge: As in the case of $$\sum 1/n^2$$, the individual terms get small very quickly, so that the sum of all of them stays finite, or, as in the case of $$\sum (-1)^{n-1}/n$$, the terms do not get small fast enough ($$\sum 1/n$$ diverges), but a mixture of positive and negative terms provides enough cancellation to keep the sum finite. You might guess from what we've seen that if the terms get small fast enough to do the job, then whether or not some terms are negative and some positive the series converges.

Theorem 11.6.1

If $$\sum_{n=0}^\infty |a_n|$$ converges, then $$\sum_{n=0}^\infty a_n$$ converges.

Proof.
Note that $$0\le a_n+|a_n|\le 2|a_n|$$ so by the comparison test $$\sum_{n=0}^\infty (a_n+|a_n|)$$ converges. Now $\sum_{n=0}^\infty (a_n+|a_n|) -\sum_{n=0}^\infty |a_n| = \sum_{n=0}^\infty a_n+|a_n|-|a_n| = \sum_{n=0}^\infty a_n$ converges by theorem 11.2.2.

$$\square$$

So given a series $$\sum a_n$$ with both positive and negative terms, you should first ask whether $$\sum |a_n|$$ converges. This may be an easier question to answer, because we have tests that apply specifically to terms with non-negative terms. If $$\sum |a_n|$$ converges then you know that $$\sum a_n$$ converges as well. If $$\sum |a_n|$$ diverges then it still may be true that $$\sum a_n$$ converges---you will have to do more work to decide the question. Another way to think of this result is: it is (potentially) easier for $$\sum a_n$$ to converge than for $$\sum |a_n|$$ to converge, because the latter series cannot take advantage of cancellation.

If $$\sum |a_n|$$ converges we say that $$\sum a_n$$ converges absolutely; to say that $$\sum a_n$$ converges absolutely is to say that any cancellation that happens to come along is not really needed, as the terms already get small so fast that convergence is guaranteed by that alone. If $$\sum a_n$$ converges but $$\sum |a_n|$$ does not, we say that $$\sum a_n$$ converges conditionally. For example $$\sum_{n=1}^\infty (-1)^{n-1} {1\over n^2}$$ converges absolutely, while $$\sum_{n=1}^\infty (-1)^{n-1} {1\over n}$$ converges conditionally.

Example 11.6.2

Does

$\sum_{n=2}^\infty {\sin n\over n^2}$

converge?

Solution

In example 11.5.2 we saw that

$\sum_{n=2}^\infty {|\sin n|\over n^2}$

converges, so the given series converges absolutely.

Example 11.6.3

Does $$\sum_{n=0}^\infty (-1)^{n}{3n+4\over 2n^2+3n+5}$$ converge?

Solution

Taking the absolute value,

$\sum_{n=0}^\infty {3n+4\over 2n^2+3n+5}$

diverges by comparison to

$\sum_{n=1}^\infty {3\over 10n},$

so if the series converges it does so conditionally. It is true that

$\lim_{n\to\infty}(3n+4)/(2n^2+3n+5)=0,$

so to apply the alternating series test we need to know whether the terms are decreasing. If we let

$f(x)=(3x+4)/(2x^2+3x+5)$

then

$f'(x)=-(6x^2+16x-3)/(2x^2+3x+5)^2,$

and it is not hard to see that this is negative for $$x\ge1$$, so the series is decreasing and by the alternating series test it converges.

### Contributors

• Integrated by Justin Marshall.

Does the series $$\sum_{n=0}^\infty {n^5\over 5^n}$$ converge? It is possible, but a bit unpleasant, to approach this with the integral test or the comparison test, but there is an easier way. Consider what happens as we move from one term to the next in this series:

$$\cdots+{n^5\over5^n}+{(n+1)^5\over 5^{n+1}}+\cdots$$

The denominator goes up by a factor of 5, $$5^{n+1}=5\cdot5^n$$, but the numerator goes up by much less:

$$(n+1)^5=n^5+5n^4+10n^3+10n^2+5n+1,$$

which is much less than $$5n^5$$ when $$n$$ is large, because $$5n^4$$ is much less than $$n^5$$. So we might guess that in the long run it begins to look as if each term is $$1/5$$ of the previous term. We have seen series that behave like this:

$$\sum_{n=0}^\infty {1\over 5^n} = {5\over4},$$

a geometric series. So we might try comparing the given series to some variation of this geometric series. This is possible, but a bit messy. We can in effect do the same thing, but bypass most of the unpleasant work.

The key is to notice that

$$\lim_{n\to\infty} {a_{n+1}\over a_n}= \lim_{n\to\infty} {(n+1)^5\over 5^{n+1}}{5^n\over n^5}= \lim_{n\to\infty} {(n+1)^5\over n^5}{1\over 5}=1\cdot {1\over5} ={1\over 5}.$$

This is really just what we noticed above, done a bit more officially: in the long run, each term is one fifth of the previous term. Now pick some number between $$1/5$$ and $$1$$, say $$1/2$$. Because

$$\lim_{n\to\infty} {a_{n+1}\over a_n}={1\over5},$$

then when $$n$$ is big enough, say $$n\ge N$$ for some $$N$$,

$${a_{n+1}\over a_n} < {1\over2} \quad \hbox{and}\quad a_{n+1} < {a_n\over2}.$$

So $$a_{N+1} < a_N/2$$, $$a_{N+2} < a_{N+1}/2 < a_N/4$$, $$a_{N+3} < a_{N+2}/2 < a_{N+1}/4 < a_N/8$$, and so on. The general form is $$a_{N+k} < a_N/2^k$$. So if we look at the series

$$\sum_{k=0}^\infty a_{N+k}= a_N+a_{N+1}+a_{N+2}+a_{N+3}+\cdots+a_{N+k}+\cdots,$$

its terms are less than or equal to the terms of the sequence

$$a_N+{a_N\over2}+{a_N\over4}+{a_N\over8}+\cdots+{a_N\over2^k}+\cdots= \sum_{k=0}^\infty {a_N\over 2^k} = 2a_N.$$

So by the comparison test, $$\sum_{k=0}^\infty a_{N+k}$$ converges, and this means that $$\sum_{n=0}^\infty a_{n}$$ converges, since we've just added the fixed number $$a_0+a_1+\cdots+a_{N-1}$$.

Under what circumstances could we do this? What was crucial was that the limit of $$a_{n+1}/a_n$$, say $$L$$, was less than 1 so that we could pick a value $$r$$ so that $$L < r < 1$$. The fact that $$L < r$$ ($$1/5 < 1/2$$ in our example) means that we can compare the series $$\sum a_n$$ to $$\sum r^n$$, and the fact that $$r < 1$$ guarantees that $$\sum r^n$$ converges. That's really all that is required to make the argument work. We also made use of the fact that the terms of the series were positive; in general we simply consider the absolute values of the terms and we end up testing for absolute convergence.

Theroem 11.7.1: The Ratio Test

Suppose that

$$\lim_{n\to \infty} |a_{n+1}/a_n|=L.$$

If

$$L < 1$$

the series $$\sum a_n$$ converges absolutely, if $$L>1$$ the series diverges, and if $$L=1$$ this test gives no information.

Proof.
The example above essentially proves the first part of this, if we simply replace $$1/5$$ by $$L$$ and $$1/2$$ by $$r$$. Suppose that $$L>1$$, and pick $$r$$ so that $$1 < r < L$$. Then for $$n\ge N$$, for some $$N$$, $${|a_{n+1}|\over |a_n|} > r \quad \hbox{and}\quad |a_{n+1}| > r|a_n|.$$ This implies that $$|a_{N+k}|>r^k|a_N|$$, but since $$r>1$$ this means that $$\lim_{k\to\infty}|a_{N+k}|\not=0$$, which means also that $$\lim_{n\to\infty}a_n\not=0$$. By the divergence test, the series diverges.

$$\square$$

To see that we get no information when $$L=1$$, we need to exhibit two series with $$L=1$$, one that converges and one that diverges. It is easy to see that $$\sum 1/n^2$$ and $$\sum 1/n$$ do the job.

Example 11.7.2

The ratio test is particularly useful for series involving the factorial function. Consider $$\sum_{n=0}^\infty 5^n/n!$$. $$\lim_{n\to\infty} {5^{n+1}\over (n+1)!}{n!\over 5^n}= \lim_{n\to\infty} {5^{n+1}\over 5^n}{n!\over (n+1)!}= \lim_{n\to\infty} {5}{1\over (n+1)}=0.$$ Since $$0 < 1$$, the series converges.

A similar argument, which we will not do, justifies a similar test that is occasionally easier to apply.

Theroem 11.7.3: The Root Test

Suppose that

$\lim_{n\to \infty} |a_n|^{1/n}=L.$

If $$L < 1$$,the series $$\sum a_n$$ converges absolutely, if $$L>1$$ the series diverges, and if $$L=1$$ this test gives no information.

The proof of the root test is actually easier than that of the ratio test, and is a good exercise.

Example 11.7.4

Analyze $$\sum_{n=0}^\infty {5^n\over n^n}$$.

Solution

The ratio test turns out to be a bit difficult on this series (try it). Using the root test:

$$\lim_{n\to\infty} \left({5^n\over n^n}\right)^{1/n}= \lim_{n\to\infty} {(5^n)^{1/n}\over (n^n)^{1/n}}= \lim_{n\to\infty} {5\over n}=0.$$

Since $$0 < 1$$, the series converges.

The root test is frequently useful when $$n$$ appears as an exponent in the general term of the series.

### Contributors

• Integrated by Justin Marshall.