Question: I cant find any clue, but I hope if I am able to show the sequence is monotone decreasing and bdd below then we are done. 
**I can see the sequence us monotone decreasing, the only disturbing part which I can't is how to bdd below.**
*I've also checked that $f(1)=-1$, $f(2)>-1$ and so on.
But I'm unable to show for all $n$.*
Please help me to solve this. 
Thanks in advance.

Answer: 
Note that

\begin{align\*}
f(n)
&= \sum\_{k=1}^{n} \left( \frac{1}{\sqrt{k}} - 2(\sqrt{k} - \sqrt{k-1}) \right) \\
&= \sum\_{k=1}^{n} \left( \frac{1}{\sqrt{k}} - \frac{2}{\sqrt{k} + \sqrt{k-1}} \right) \\
&= \sum\_{k=1}^{n} \frac{\sqrt{k-1} - \sqrt{k}}{\sqrt{k}(\sqrt{k} + \sqrt{k-1})} \\
&= - \sum\_{k=1}^{n} \frac{1}{\sqrt{k}(\sqrt{k} + \sqrt{k-1})^2}
\end{align\*}

I hope this is enough for you to conclude the convergence of $f(n)$. Also, if we are allowed to use a bit of calculus, then we can come up with a more systematic solution rather than this tricky approach. (Which will be essentially a toy version of Euler-MacLaurin expansion.)
HINT.-$f(x+h)-f(x)=2\sqrt x+\dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{x+2}}+\cdots+\dfrac{1}{\sqrt{x+h}}-2\sqrt{x+h}$.

It is easy enough to prove that $\lim\_{x\to\infty}(2\sqrt x-2\sqrt {x+h})=0$.

Then $\{f(n)\}\_{n\in\mathbb N}$ is a Cauchy sequence so it have a limit in $\mathbb R$.