Convergence of a sum, which test should I use? I have to show if this sum
$$\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}$$
converges or not.
I tried to do something like
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$$
but I got nothing, because the best I could make it was:
$$\frac{\sqrt{n+2}-\sqrt{n+1}}{\sqrt{n+2}}\cdot\frac{1}{\sqrt{n+1}}\cdot\frac{\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}\cdot\frac{\sqrt{n}}{1}$$
And I really don't know how to simplify this expression.
 A: It is convergent by comparing with $1/n^2$.
A: As others have noted, convergence can be shown by comparing the series with the (convergent) series $\sum {1\over n^2}$.  Here's one slick way to obtain the comparison:
$$\begin{align}
0&\lt\left({1\over\sqrt n}-{1\over\sqrt{n+1}}\right){1\over\sqrt n}\\\\
&\lt\left({1\over\sqrt n}-{1\over\sqrt{n+1}}\right)\left({1\over\sqrt n}+{1\over\sqrt{n+1}}\right)\\\\
&={1\over n}-{1\over n+1}\\\\
&={1\over n(n+1)}\\\\
&\lt{1\over n^2}
\end{align}$$
A: Three methods, increasingly involved (but increasingly generalizable):
To use a very simple trick:
As for any $n\geq 1$ we have 
$$
0 \leq \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}
\leq \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
$$
we can immediately conclude by comparison, since the series $\sum_n \left( \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$ converges (it is a telescopic series):
$$
\sum_{n=1}^N \left( \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}} \right)
= 1 - \frac{1}{\sqrt{N+1}} \xrightarrow[N\to\infty]{} 1.
$$

To use only simple manipulations:
$$\begin{align}
0 \leq \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}
&= 
\left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}
= 
\left(\frac{\sqrt{1+\frac{1}{n}}-1}{\sqrt{n}\sqrt{n+1}}\right)\\
&= 
\frac{1}{n}\left(\frac{\sqrt{1+\frac{1}{n}}-1}{\sqrt{1+\frac{1}{n}}}\right)\\
&\leq 
\frac{1}{n}\left( \sqrt{1+\frac{1}{n}}-1 \right)\\
&=\frac{1}{n}\left( \frac{1+\frac{1}{n}-1}{\sqrt{1+\frac{1}{n}}+1} \right)
=\frac{1}{n^2}\left( \frac{1}{\sqrt{1+\frac{1}{n}}+1} \right)\\
&\leq \frac{1}{2n^2}
\end{align}$$
and conclude by comparison.

To use Taylor series: since $\sqrt{1+u} = 1+\frac{u}{2} + o(u)$ and $\frac{1}{1+u} = 1-u + o(u)$ when $u\to0$,
$$
\sqrt{n+1} = \sqrt{n}\sqrt{1+\frac{1}{n}} = \sqrt{n}\left(1+\frac{1}{2n}+o\left(\frac{1}{n}\right)\right)
$$
so
$$\begin{align}
\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} &= \frac{1}{\sqrt{n}}\left( 1-\frac{1}{1+\frac{1}{2n}+o\left(\frac{1}{n}\right)} \right)
= \frac{1}{\sqrt{n}}\left( 1-\left(1-\frac{1}{2n}+o\left(\frac{1}{n}\right)\right) \right)\\
&= \frac{1}{\sqrt{n}}\left( \frac{1}{2n}+o\left(\frac{1}{n}\right) \right) \\
&= \frac{1}{2n^{3/2}}+o\left(\frac{1}{n^{3/2}}\right)
\end{align}$$
and finally
$$
\left(\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}\right)\frac{1}{\sqrt{n}}
= \frac{1}{2n^{2}}+o\left(\frac{1}{n^{2}}\right)
$$
and you can again conclude by comparison.
