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

  • 2
    $\begingroup$ This is not an alternating series, so the test does not apply. Use instead the comparison test. with $\sum{1\over\sqrt{2}^n}$ $\endgroup$ Jul 30, 2014 at 3:09
  • $\begingroup$ Do you mean $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{n}{2^n}$? $\endgroup$
    – David
    Jul 30, 2014 at 3:11
  • $\begingroup$ Yes that's what I meant. Thanks I just fixed it. Sorry. $\endgroup$
    – Oyukyfairy
    Jul 30, 2014 at 3:12
  • $\begingroup$ Don't worry specifically about the $n=1$ case. Try to prove it for all $n \geq 1$. $\endgroup$
    – foobar1209
    Jul 30, 2014 at 3:21
  • $\begingroup$ Thanks everyone! I am trying to get through Calc II and it isn't fun...and I have an exam on this stuff tomorrow. $\endgroup$
    – Oyukyfairy
    Jul 30, 2014 at 3:35

4 Answers 4


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


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.

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


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