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.
It is convergent by comparing with $1/n^2$.
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.
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:
$$\begin{align} 0&\lt\left({1\over\sqrt n}-{1\over\sqrt{n+1}}\right){1\over\sqrt n}\\\\ &\lt\left({1\over\sqrt n}-{1\over\sqrt{n+1}}\right)\left({1\over\sqrt n}+{1\over\sqrt{n+1}}\right)\\\\ &={1\over n}-{1\over n+1}\\\\ &={1\over n(n+1)}\\\\ &\lt{1\over n^2} \end{align}$$
