Q:

Calculate limit of $\frac{1}{n}\cdot (1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})$

How to prove that
$$\lim_{n\to\infty}(\frac{1}{n}\cdot(1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})) = +\infty$$
using only basic limit operations and theorems?

A:

Using the Stolz-Cesaro Theorem,
$$\lim_{n\to\infty}\frac{1+\sqrt{2}+...+\sqrt{n}}{n}=\lim_{n\to\infty} \sqrt{n}=\infty$$

A:

You have
$$\begin{align}
\frac{1}{n}\sum_{k=1}^n \sqrt{k}
&\geq \frac{1}{n}\sum_{k=\lfloor n/2\rfloor}^n \sqrt{k}
\geq \frac{1}{n}\sum_{k=\lfloor n/2\rfloor}^n \sqrt{\frac{n}{2}} \\
&= \frac{n-\lfloor n/2\rfloor+1}{n}\cdot \sqrt{\frac{n}{2}}
\geq \frac{n}{2n}\cdot \sqrt{\frac{n}{2}} \\
&= \frac{\sqrt{n}}{2\sqrt{2}} \xrightarrow[n\to\infty]{} \infty
\end{align}$$
where the only "trick" lies in the first inequality: dropping some (positive) terms from the sum actually helps to get an easy lower bound on the sum.

