Does $\sum_{n=1}^{\infty} (\frac{1}{n}-\frac{1}{\sqrt{n(n+1)}})$ converge $$\sum_{n=1}^{\infty} \left ( \frac{1}{n}-\frac{1}{\sqrt{n(n+1)}}\right )=\sum_{n=1}^{\infty} \left ( \frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}}\right )$$
By ratio test I get $$ \lim_{n\to \infty} \frac{\frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}}}{\frac{1}{n}}=\frac{0}{0}$$
So i can't conclude anything. I tried comparison tests, but i get $\le$ something divergent.
 A: \begin{align*}
\sum_{n=1}^{\infty}\frac{\sqrt{n(n+1)}-n}{n\sqrt{n(n+1)}}
&=\sum_{n=1}^{\infty}\frac{(\sqrt{n(n+1)}-n)(\sqrt{n(n+1)}+n)}{n\sqrt{n(n+1)}(\sqrt{n(n+1)}+n)}\\
&=\sum_{n=1}^{\infty}\frac1{\sqrt{n(n+1)}(\sqrt{n(n+1)}+n)}\\
&=\sum_{n=1}^{\infty}\frac1{n(n+1)+n\sqrt{n(n+1)}}\\
&\le\sum_{n=1}^{\infty}\frac1{n^2}.
\end{align*}
A: Use Taylor's expansion at order $1$:
\begin{align*}\frac1n-\frac1{\sqrt{n(n+1)}}&=\frac1n\Biggl(1-\frac1{\sqrt{1+\frac1n}}\Biggr)=\frac1n\biggl(1-1+\frac1{2n}+o\Bigl(\frac1n\Bigr)\biggr)\\ &=\frac1{2n^2}+o\biggl(\frac1{n^2}\biggr)\sim_\infty\frac1{2n^2} \end{align*}
and the latter converges.
A: Compare to $$\sum_{n=1}^\infty\left(\frac1n-\frac1{\sqrt{n\cdot n}}\right)$$ on one side and $$\sum_{n=1}^\infty\left(\frac1n-\frac1{\sqrt{(n+1)(n+1)}}\right)$$on the other.
A: You don't need ratio test or a clever bound for the general term... Like most problems on MSE, a simple asymptotic expansion fits the bill.
$$\frac{1}{n}-\frac{1}{\sqrt{n(n+1)}} = \frac{1}{2n^2} + o\left(\frac{1}{n^2}\right)$$
