How to prove the series ${S_n}$ converges where $S_n$ is
$$\sum_{k=1}^n\frac{1}{\sqrt{k\cdot n}}$$
and what the limit is?
I thought it was $p$-series where $p=1/2<1$ so it must diverge but according to the question, it must converge, I'm not sure about which convergence test I can use.
Any tip is much appreciated
By Stolz-Cesàro theorem ($*/\infty$ case): \begin{align} & \lim_{n \to \infty}\frac{\sum_{k = 1}^n\frac{1}{\sqrt{k}}}{\sqrt{n}} = \lim_{n \to \infty}\frac{\frac{1}{\sqrt{n + 1}}}{\sqrt{n + 1} - \sqrt{n}} = \lim_{n \to \infty}\frac{1}{\sqrt{n + 1}}(\sqrt{n + 1} + \sqrt{n}) \\ =& 1 + \lim_{n \to \infty}\sqrt{\frac{n}{n + 1}} = 2. \end{align}
