Limit of a sum $\frac{1}{\sqrt n}$ $$
\lim_{n \longrightarrow \infty} \frac{1}{\sqrt1}+ \frac{1}{\sqrt 2}+...+\frac{1}{\sqrt n}
$$
I have been stuck with this limit for some time and need some help to keep going. I can see it won't converge but I obviously have to prove it. I tried to rewrite it as:
$$
\lim_{n \longrightarrow \infty} \frac{1+\sqrt 2}{\sqrt 2}+ \frac{\sqrt 3 + \sqrt 4}{\sqrt {12}}+...+\frac{\sqrt n + \sqrt {n-1}}{\sqrt {n(n-1)}},
$$
but it wasn't very useful.
I also tried to approach it as a Riemann sum, but I didn't really see the light either because I couldn't figure out anything more than
$$
\lim _{n \longrightarrow \infty}\sum ^n _{i=1} \frac{1}{\sqrt i}
$$
I would really appreciate any help.
 A: The sum has $n$ terms and each term is bounded from below by $\frac{1}{\sqrt{n}}$, so the sum is bounded from below by $\frac{n}{\sqrt{n}} = \sqrt{n}$. Of course, $\lim_{n \to \infty} \sqrt{n} = +\infty$.
(Using integrals or sophisticated criteria is an overkill in this case.)
A: Here is another way to approach it differently from the comparison test.
More precisely, you can apply the condensation test.
If $a_{n}$ is non-negative and decreasing, then the following relation holds:
\begin{align*}
\sum_{n=1}^{\infty}a_{n} \ \ \text{converges} \ \ \text{iff} \ \ \sum_{n=1}^{\infty}2^{n}a_{2^{n}} \ \ \text{converges}
\end{align*}
Generically speaking, the following series
\begin{align*}
\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}} = 1 + \frac{1}{2^{\alpha}} + \frac{1}{3^{\alpha}} + \ldots
\end{align*}
converges iff the following series
\begin{align*}
\sum_{n=1}^{\infty}2^{n}\times\frac{1}{(2^{n})^{\alpha}} = \sum_{n=1}^{\infty}[2^{(1 - \alpha)}]^{n}
\end{align*}
converges, which is a geometric series with ratio $r = 2^{1 - \alpha}$.
As it is known, a geometric series converges iff its ratio satisfies $|r| < 1$.
Consequently, the proposed series converges iff
\begin{align*}
2^{1 - \alpha} < 1 & \Longleftrightarrow 2^{1 - \alpha} < 2^{0}\\\\
& \Longleftrightarrow 1 - \alpha < 0\\\\
& \Longleftrightarrow \alpha > 1
\end{align*}
and diverges otherwise. Hence we conclude the proposed series diverges since $\alpha = 1/2$.
Hopefully this helps !
