Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to know if the sum $$\sum_{n=1}^\infty(-1)^n\frac{\sqrt{n+1}-\sqrt{n}}{n}$$ converges or not. So I've tried the alternating series text:

  1. $\lim\limits_{n\rightarrow\infty}\frac{\sqrt{n+1}-\sqrt{n}}{n}=0$ is clear.
  2. I also need $a_{n+1}\leq a_n$.

So my question is if there is an very easy way to show 2. ? (I've tried to calculate the inequation but I don't get a nice result.)

share|cite|improve this question
up vote 13 down vote accepted



Clearly, $a_n$ is decreasing.

share|cite|improve this answer
Which also shows that this sum absolutely converges. – Amihai Zivan Jan 1 '13 at 14:39
so I have $\frac{1}{(n+1)(\sqrt{n+2}+\sqrt{n+1})}\leq\frac{1}{n(\sqrt{n+1}+\sqrt{n})}$ but why does it help me? – user45389 Jan 1 '13 at 14:56
This shows that $a_n$ is decreasing – Amr Jan 1 '13 at 14:58
sry but I don't see it :( – user45389 Jan 1 '13 at 15:00
You wanted to show that $a_n\geq a_{n+1}$. Now you are done, because $a_n=\frac{1}{n(n^{0.5}+(n+1)^{0.5})}$ is decreasing – Amr Jan 1 '13 at 15:01

The answer by @Amr is very nice and solves your problem. Here are some other approaches.

I. Let $a_n = (\sqrt{n+1}-\sqrt{n})/n$. Then $$\begin{eqnarray*} \frac{a_{n+1}}{a_n} &=& \frac{\sqrt{n+2}-\sqrt{n+1}}{n+1} \frac{n}{\sqrt{n+1}-\sqrt{n}} \\ &=& \frac{n}{n+1}(\sqrt{n+2}-\sqrt{n+1})(\sqrt{n}+\sqrt{n+1}) \\ &<& \frac{n}{n+1}(\sqrt{n+2}-\sqrt{n+1})(\sqrt{n+2}+\sqrt{n+1}) \\ &=& \frac{n}{n+1}. \end{eqnarray*}$$

II. Since $$\frac{a_{n+1}}{a_n} \sim 1 - \frac{3}{2n} + \ldots \qquad (n\to\infty)$$ the series converges absolutely by Raabe's test.

share|cite|improve this answer

Another way is to define a function $f$ where $f(x)=\displaystyle\frac{\sqrt{x+1}-\sqrt{x}}{x}$ you get

$f'(x)=\dfrac{\sqrt{{x}^{2}+x}-x-2}{2{x}^{2}{\cdot}\sqrt{x+1}}$ which is strictly negative since $\sqrt{x^2+x}<x+2$ hence $f$ is decreasing and so is $a_n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.