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.)
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> 
> With a [Riemann Zeta Function Identity](https://dlmf.nist.gov/25.2.E8):
> 
> 
> 

\begin{align}
\mrm{f}\pars{n} & \equiv 1 + 1/\root{2} + 1/\root{3} + \cdots +
1/\root{n} - 2\root{n} =
\sum\_{k = 1}^{n}{1 \over k^{\color{red}{1/2}}} - 2\root{n}
\\[5mm] & =
\bracks{\zeta\pars{\color{red}{1 \over 2}} -
{n^{1 - \color{red}{1/2}} \over \color{red}{1/2} - 1} +
\color{red}{1 \over 2}\int\_{n}^{\infty}{x - \left\lfloor\,{x}\,\right\rfloor \over x^{\color{red}{1/2} + 1}}\,\dd x} - 2\root{n}
\\[5mm] & =
\zeta\pars{1 \over 2} +
{1 \over 2}\int\_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x
\end{align}

> 
> Note that
>  $\ds{0 <
> {1 \over 2}\int\_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x <
> {1 \over 2}\int\_{n}^{\infty}{\dd x \over x^{3/2}} = {1 \over \root{n}}
> \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\LARGE\to}\,\,\,\color{red}{\large 0}}$
> 
> 
> 

$$
\bbx{\mbox{Then,}\quad\lim\_{n \to \infty}\mrm{f}\pars{n} = \zeta\pars{1 \over 2}} \approx -1.4604 
$$