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2.3 Convergence tests

We now collect a few test procedures that can be used to find out whether a series converges or diverges. However, not all series can be treated with one of these tests, and sometimes it is necessary to find out “manually” as we shall see.

Theorem 2.3.1 (The simple comparison test).

Let $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ be two series. Suppose there is $N\in\mathbb{N}$ such that $0\leqslant a_{n}\leqslant b_{n}$ , for all $n>N$ . If $\sum_{n=1}^{\infty}b_{n}$ is convergent then $\sum_{n=1}^{\infty}a_{n}$ is convergent as well. In other words, if $\sum_{n=1}^{\infty}a_{n}$ is divergent then $\sum_{n=1}^{\infty}b_{n}$ is divergent as well.

Comment: This is useful if you can compare “term-by-term” with a sequence for which you already know whether it converges or diverges.

Proof.

To be done in MATH210. ∎

Example 2.3.2.

(1)

The series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{n}+1}$ is convergent, because $\displaystyle 0\leq\frac{1}{2^{n}+1}<\frac{1}{2^{n}}$ for all $n$ and $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{n}}$ is convergent.
(2)

The series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ is convergent, because $\displaystyle 0\leq\frac{1}{n^{2}}\leqslant\frac{2}{n(n+1)}$ (note that $n(n+1)=n^{2}+n\leq 2n^{2}$ ) and the series $\displaystyle\sum_{n=1}^{\infty}\frac{2}{n(n+1)}$ is convergent by Proposition 2.2.3.
(3)

The series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}$ is convergent if $\alpha\geq 2$ and divergent if $\alpha\leq 1$ . If $\alpha\geq 2$ then $\displaystyle\frac{1}{n^{\alpha}}\leq\frac{1}{n^{2}}$ for all $n\geq 1$ , and $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ is convergent. If $\alpha\leq 1$ then $\displaystyle\frac{1}{n^{\alpha}}\geq\frac{1}{n}$ for all $n\geq 1$ , and $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. It can actually be shown to be convergent for all $\alpha>1$ .

Theorem 2.3.3 (The limit comparison test).

Let $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ be two series. Suppose there is $N\in\mathbb{N}$ such that $a_{n}>0$ and $b_{n}>0$ for all $n>N$ , and that $a_{n}/b_{n}$ converges to a non-zero limit as $n\to\infty$ . Then $\sum_{n=1}^{\infty}a_{n}$ is convergent if and only if $\sum_{n=1}^{\infty}b_{n}$ is convergent.

Comment: This is useful if you can compare “asymptotically” with a sequence for which you already know whether it converges or diverges.

Proof.

To be done in MATH210. ∎

Example 2.3.4.

Is $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3n+4}$ convergent or divergent?

To answer this question, we can use the simple comparison test as well as the limit comparison test.

By the limit comparison test: if $\displaystyle a_{n}=\frac{1}{3n+4}$ and $\displaystyle b_{n}=\frac{1}{n}$ then $a_{n}/b_{n}\to 1/3$ as $n\to\infty$ ; since $\sum_{n=1}^{\infty}b_{n}$ is divergent, so is $\sum_{n=1}^{\infty}a_{n}$ .

By the simple comparison test: since $3n+4\leq 4n$ whenever $n\geq 4$ , and $\displaystyle\sum_{n=1}^{\infty}\frac{1}{4n}$ is divergent, so is $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3n+4}$ .

Example 2.3.5.

Let us instead look at the series

\sum_{n=1}^{\infty}\frac{3}{n^{2}}.

We know from Example 2.3.2(2) that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ converges, but $\displaystyle a_{n}=\frac{3}{n^{2}}>\frac{1}{n^{2}}$ for all $n\in\mathbb{N}$ , so we cannot apply the simple comparison test. Instead $a_{n}/b_{n}=3\rightarrow 3\not=0$ as $n\rightarrow\infty$ , so we can apply the limit comparison test to show that $\displaystyle\sum_{n=1}^{\infty}\frac{3}{n^{2}}$ converges.

Proposition 2.3.6 (The ratio test).

Let $\sum_{n=1}^{\infty}a_{n}$ be a series. Suppose there is $N\in\mathbb{N}$ such that $a_{n}>0$ for all $n>N$ , and suppose there is $\ell\in\mathbb{R}$ such that $a_{n+1}/a_{n}\to\ell$ as $n\to\infty$ . Then

•

if $\ell<1$ then $\sum_{n=1}^{\infty}a_{n}$ converges;
•

if $\ell>1$ then $\sum_{n=1}^{\infty}a_{n}$ diverges;
•

if $\ell=1$ then $\sum_{n=1}^{\infty}a_{n}$ may converge or diverge.

Comment: The top candidate among the tests – this is useful in many cases, especially if you cannot compare with another sequence but instead you know something about the “asymptotic behaviour” of the ratios $a_{n+1}/a_{n}$ .

Proof.

To be done in MATH210. ∎

Example 2.3.7.

Let $a_{n}=\displaystyle\frac{n}{2^{n}}$ . Then

\frac{a_{n+1}}{a_{n}}=\frac{n+1}{2^{n+1}}\,\frac{2^{n}}{n}=\frac{n+1}{2n}\to% \frac{1}{2}<1\qquad\mbox{as}\quad n\to\infty,

so $\sum_{n=1}^{\infty}a_{n}$ is convergent by the ratio test.

When a series contains both positive and negative terms, the sequence $(A_{n})_{n\in\mathbb{N}}$ of partial sums is no longer increasing, and there are various ways in which it might fail to converge. For example, if $a_{n}=(-1)^{n-1}$ then the partial sums alternate between $1$ and $0$ .

Proposition 2.3.8.

If $\sum_{n=1}^{\infty}|a_{n}|$ is convergent then so is $\sum_{n=1}^{\infty}a_{n}$ . Furthermore, $|\sum_{n=1}^{\infty}a_{n}|\leqslant\sum_{n=1}^{\infty}|a_{n}|$ .

Comment: Good if you have a series with some positive and some negative terms and you know that without the minus signs it would already converge.

Proof.

Exercise X. ∎

Definition 2.3.9.

The series $\sum_{n=1}^{\infty}a_{n}$ is called absolutely convergent if the series $\sum_{n=1}^{\infty}|a_{n}|$ converges.

Example 2.3.10.

Consider the series $\displaystyle\sum_{n=1}^{\infty}\sin(n)\frac{1}{n^{2}}$ . We know that $\displaystyle 0\leq|\sin(n)\frac{1}{n^{2}}|\leq\frac{1}{n^{2}}$ , for all $n\in\mathbb{N}$ , and the series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}$ converges according to Example 2.3.2(2). By the simple comparison test, the series $\displaystyle\sum_{n=1}^{\infty}|\sin(n)\frac{1}{n^{2}}|$ must converge, and by Proposition 2.3.8, $\displaystyle\sum_{n=1}^{\infty}\sin(n)\frac{1}{n^{2}}$ therefore converges.

Exercise W3.1.

Given a sequence $(a_{n})_{n\in\mathbb{N}}$ .

(1)

If $a_{n}\in\mathbb{R}$ , does convergence of $\sum_{n=1}^{\infty}a_{n}$ imply absolute convergence? Does absolute convergence imply convergence? Explain briefly. If you have more time, then prove Proposition 2.3.8.
(2)

Suppose now that all $a_{n}$ are negative numbers. What can you say about (1) in this case? Explain briefly.

Theorem 2.3.11 (Leibniz’s criterion on alternating series).

If $(a_{n})_{n\in\mathbb{N}}$ is decreasing and tends to $0$ then $\sum_{n=1}^{\infty}(-1)^{n-1}a_{n}$ is convergent.

Comment: Applicable if you have alternatingly positive and negative consecutive terms in your series.

Proof.

Let $A_{2n}=(a_{1}-a_{2})+(a_{3}-a_{4})+\cdots+(a_{2n-1}-a_{2n})$ . Then the sequence $(A_{2n})_{n\in\mathbb{N}}$ is increasing, since

A_{2n}-A_{2n-2}=a_{2n-1}-a_{2n}\geqslant 0,\quad n\in\mathbb{N}.

Also, $(A_{2n})_{n\in\mathbb{N}}$ is bounded above by $a_{1}$ , since

A_{2n}=a_{1}-(a_{2}-a_{3})-(a_{4}-a_{5})-\cdots-(a_{2n-2}-a_{2n-1})-a_{2n}\leq a% ,\quad n\in\mathbb{N}.

As we remember from MATH113, a bounded and increasing sequence converges, so $(A_{2n})_{n\in\mathbb{N}}$ has a limit $A\in\mathbb{R}$ .

Since $a_{n}\to 0$ as $n\rightarrow\infty$ and the sum of convergent sequences is convergent, we see that the sequence $(A_{2n-1})_{n\in\mathbb{N}}=(A_{2n}-a_{2n})_{n\in\mathbb{N}}$ converges to $A-0=A$ , too. Hence the whole sequence $(A_{n})_{n\in\mathbb{N}}$ (with both even and odd $n$ ) converges, with limit $A$ . ∎

Key idea of proof.

prove first convergence for even terms using monotonicity, then use $a_{n}\rightarrow 0$

Example 2.3.12.

Let $\displaystyle a_{n}=\frac{(-1)^{n}}{n}$ . By Theorem 2.3.11, $\sum_{n=1}^{\infty}a_{n}$ is convergent because $(|a_{n}|)_{n\in\mathbb{N}}$ is decreasing and tends to $0$ . However, $\sum_{n=1}^{\infty}|a_{n}|$ is divergent according to Proposition 2.2.4 since $\displaystyle|a_{n}|=\frac{1}{n}$ .

Exercise W3.2.

Consider the series $\sum_{n=1}^{\infty}a_{n}$ defined by

a_{n}=\left\{\begin{array}[]{l@{\; : \;}l}\frac{1}{n}\hfil\;:\;&n\leq 100\\ (-1)^{n}\frac{1}{n}\hfil\;:\;&n>100.\end{array}\right.

Does the series converge or diverge?

Example 2.3.13.

(1)

Consider the series

$\sum_{n=1}^{\infty}\frac{(-1)^{n}n}{n+1}.$

Since $\displaystyle\frac{n}{n+1}\rightarrow 1$ as $n\rightarrow\infty$ , we cannot apply Leibniz’ criterion. In fact, let $A_{n}$ denote the $n$ -th partial sum for the series. Suppose $A$ were the limit of $(A_{n})_{n\in\mathbb{N}}$ and consider $\varepsilon=\frac{1}{4}$ . Then since

$|A_{n+1}-A_{n}|=\frac{n+1}{n+2}>\frac{1}{2},\quad n\in\mathbb{N},$

we see that, for all $n\in\mathbb{N}$ , either $|A_{n+1}-A|>\frac{1}{4}$ or $|A_{n}-A|>\frac{1}{4}$ . This contradicts $A$ being the limit of $(A_{n})_{n\in\mathbb{N}}$ , so the sequence of partial sums does not converge and hence the series $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}n}{n+1}$ is divergent.
(2)

Consider the following series

$\sum_{n=1}^{\infty}n^{2}\operatorname{e}^{-5n}.$

Since all terms are positive, we do not need Leibniz’ theorem. On the other hand, we do not have any suitable series to compare to. The $n^{2}$ parts seems to diverge, while the $e^{-n}$ changes the situation. A good test in general is the ratio test, which yields

$\Big{|}\frac{a_{n+1}}{a_{n}}\Big{|}=\frac{(n+1)^{2}\operatorname{e}^{-(n+1)}}{% n^{2}\operatorname{e}^{-n}}=\frac{(n+1)^{2}}{n^{2}}\operatorname{e}^{-1}% \rightarrow\operatorname{e}^{-1},\quad n\rightarrow\infty,$

Since the limit $\operatorname{e}^{-1}<1$ , the ratio test implies that the series $\displaystyle\sum_{n=1}^{\infty}n^{2}\operatorname{e}^{-5n}$ converges.
(3)

Some series of your choice: