# Prove series is convergent

I am trying to prove that the series $\sum\limits_{n=2}^\infty=\frac{\sqrt{n+1}-\sqrt{n}}{n}$ is convergent. I tried using the comparison test, but it's not showing me that the series is convergent.

These are my steps:

Using $\sum\limits_{n=2}^\infty = \frac{1}{n^{3/2}}$

$$\lim_{n\to\infty}\frac{\frac{\sqrt{n+1}-\sqrt{n}}{n}}{\frac{1}{n^{3/2}}} = \lim_{n\to\infty}(\sqrt{n})(\sqrt{n+1}-\sqrt{n})=\lim_{n\to\infty}n+\sqrt{n}-n=\sqrt{n}$$

This shows that the series in non-convergent, which I know not to be the case.

• That is a divergent series. – Jonathan Hebert Mar 12 '15 at 2:46
• Sorry, I made a typo. It was subtraction, not addition. – Richard Hum Mar 12 '15 at 2:47

Multiplying by $\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$, we get $$\frac{\sqrt{n+1}-\sqrt{n}}{n}=\frac1{(\sqrt{n+1}+\sqrt{n})n}$$ Now compare with $\sum\limits_{n=1}^\infty\frac1{n\sqrt{n}}$
hint: $\dfrac{\sqrt{n+1}-\sqrt{n}}{n} < \dfrac{1}{2n\sqrt{n}}$
Another (less elegant) solution could be $$a_n=\frac{\sqrt{n+1}-\sqrt{n}}{n}=\frac{\sqrt n\sqrt{1+\frac 1n}-\sqrt n}{n}=\frac{\sqrt{1+\frac 1n}-1}{\sqrt n}$$ Now, use the fact that, for small values of $x$, $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+O\left(x^3\right)$. Replacing $x=\frac 1n$ then gives $$a_n=\frac{1}{2} \left(\frac{1}{n}\right)^{3/2}-\frac{1}{8} \left(\frac{1}{n}\right)^{5/2}+O\left(\left(\frac{1}{n}\right)^{7/2}\right)$$ and the result given by BRICS-Fan.
• Sorry if you don't practice Taylor series; big $O$ notation describes the limiting behavior of a function when the argument tends towards a particular value. For infinitesimal asymptotics, big $O$ can also be used to describe the error term in an approximation to a mathematical function. – Claude Leibovici Mar 12 '15 at 4:01