# How to apply the alternating series test to the series $\sum (-1)^{n+1} n/2^n$?

So, I need to test the following series for convergence or divergence:

$$\sum_{n=1}^\infty (-1)^{n+1}{n\over {2^n}}$$

I know that when you use the Alternating Series Test, the series must satisfy two conditions. Which are:

1. $$b_{n+1} \le b_n$$
2. $$\lim_{n\to \infty} b_n =0$$

I having a hard time with the first condition because if I use 1 for n then I have a problem. This is my work so far:

$${(n+1)\over 2^{(n+1)}} ? {n\over 2^n}$$ $${(1+1)\over 2^{(1+1)}} ? {1\over 2^1}$$ $${(2)\over 2^{(2)}} ? {1\over 2^1}$$ $${2\over 4} = {1\over 2}$$ They end up equaling each other. On the other hand, if I plug in 2, I get something that does satisfy the first condition.$${(n+1)\over 2^{(n+1)}} ? {n\over 2^n}$$ $${(2+1)\over 2^{(2+1)}} ? {2\over 2^2}$$ $${(3)\over 2^{(3)}} ? {2\over 2^2}$$ $${3\over 8} ? {2\over 4}$$ $${3\over 8} \le {1\over 2}$$

So... what do I do? Thanks in advance

• This is not an alternating series, so the test does not apply. Use instead the comparison test. with $\sum{1\over\sqrt{2}^n}$ Jul 30, 2014 at 3:09
• Do you mean $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{n}{2^n}$? Jul 30, 2014 at 3:11
• Yes that's what I meant. Thanks I just fixed it. Sorry. Jul 30, 2014 at 3:12
• Don't worry specifically about the $n=1$ case. Try to prove it for all $n \geq 1$. Jul 30, 2014 at 3:21
• Thanks everyone! I am trying to get through Calc II and it isn't fun...and I have an exam on this stuff tomorrow. Jul 30, 2014 at 3:35

In that case, it's pretty straightforward:

First note that $b_{n}\ge b_{n+1}\iff b_n-b_{n+1}\ge 0$, then consider the reformulated question:

$$b_{n}-b_{n+1}={n+1\over 2^{n+1}}-{n\over 2^n}\stackrel{?}{\ge}0$$

this is true if and only if the same is true after multiplying both sides by $2^{n+1}$, because multiplying both sides of an inequality by a positive number keep the inequality true, so

$$2n-(n+1)=n-1\stackrel{?}{\ge}0$$

which is true so long as $n\ge 1$, so that checks out.

To see the limit goes to zero you can proceed either by using

$$\lim_{x\to\infty}{x\over 2^x}=\lim_{x\to\infty} {1\over 2^x\ln 2}=0$$

using L'Hôpital's rule or by noting $n\le \sqrt{2}^n$ for every $n\ge 1$

This is the same as saying $n^2\le 2^n$ for every $n\ge 1$, which is verified by base case: $n=1$, $1\le 2$ check. Assume it's true for some $n\ge 1$, then

$$(n+1)^2=n^2+2n+1\le 2^n +2n+1\le 2^n+n^2 < 2^n+2^n=2^{n+1}$$

by inductive hypothesis, and since $2n+1\le n^2$ for $n\ge 1$, so the rest follows by induction, and we have

$$0\le\lim_{n\to\infty}{n\over 2^n}\le\lim_{n\to\infty}{(\sqrt{2})^n\over 2^n}=\lim_{n\to\infty}{1\over(\sqrt{2})^n}=0$$

which settles that by the squeeze theorem.

• Sorry, but where did the n-1 come from? Jul 30, 2014 at 3:28
• $2n-(n+1) - 2n-n-1=n-1$ Jul 30, 2014 at 3:28
• Oh okay thanks. I get it know! Jul 30, 2014 at 3:30
• Oh alright. Thank you once again! Jul 30, 2014 at 3:33

You can use induction. The first step is obvious. Assume $\frac{k}{2^k}>\frac{k+1}{2^{k+1}}$. Then $\frac{k+1}{2^{k+1}}=\frac{1}{2}\frac{k}{2^{k}}+\frac{1}{2} \frac{1}{2^{k}}>\frac{1}{2}\frac{k+1}{2^{k+1}}+\frac{1}{2} \frac{1}{2^{k+1}}=\frac{k+2}{2^{k+2}}$.

As corrected, this is an alternating series.

Also, for the alternating series test to be applied to $\sum_{k=1}^{\infty} (-1)^n a_n$, all that is needed is that $a_{n+1} < a_n$ for all $n > N$ for some $N$. Any initial part of a series is irrelevant when discussing convergence.

In this case, the fact that $\dfrac{n}{2^n}$ is eventually decreasing has been discussed here too many times.

To check if $\dfrac{n}{2^n}$ is eventually decreasing, it suffices to show that the derivative of $\displaystyle f(x)=\frac{x}{2^x}$ is eventually negative.

To do so, note that $$f^\prime(x)=\frac{2^x-x\cdot 2^x\ln2}{2^{2x}}=\frac{1-x\cdot\ln{2}}{2^x}$$ For $x>1/\ln2$ we have $f^\prime(x)<0$. Hence $\dfrac{n}{2^n}$ is eventually decreasing.