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.)
**Approach $\bf{1}$**

One way to bound such series is to note that
$$
\int\_n^{n+1}\frac{\mathrm{d}x}{\sqrt{x}}\le\frac1{\sqrt{n}}\le\int\_{n-1}^n\frac{\mathrm{d}x}{\sqrt{x}}
$$
Adding up the integrals is simple.

---

**Approach $\bf{2}$**

Another way is to note that
$$
2\sqrt{n+1}-2\sqrt{n\vphantom{+1}}=\frac2{\sqrt{n+1}+\sqrt{n\vphantom{+1}}}
$$
That is,
$$
2\sqrt{n+1}-2\sqrt{n\vphantom{+1}}\le\frac1{\sqrt{n\vphantom{+1}}}\le2\sqrt{n\vphantom{-1}}-2\sqrt{n-1}
$$
The bounds can be summed as a telescoping series.

---

**Approach $\bf{3}$**
$$
\begin{align}
\frac1{\sqrt{n}}-2\sqrt{n\vphantom{-1}}+2\sqrt{n-1}
&=\frac1{\sqrt{n}}-\frac2{\sqrt{n\vphantom{-1}}+\sqrt{n-1}}\\
&=\frac{\sqrt{n-1}-\sqrt{n\vphantom{-1}}}{n+\sqrt{n(n-1)}}\\[6pt]
&\lt0
\end{align}
$$
This implies that the sequence
$$
a\_n=\sum\_{k=1}^n\frac1{\sqrt{k}}-2\sqrt{n}
$$
is decreasing.
$$
\begin{align}
\frac1{\sqrt{n}}-2\sqrt{n+1}+2\sqrt{n\vphantom{+1}}
&=\frac1{\sqrt{n}}-\frac2{\sqrt{n+1}+\sqrt{n\vphantom{+1}}}\\
&=\frac{\sqrt{n+1}-\sqrt{n\vphantom{+1}}}{n+\sqrt{n(n+1)}}\\[6pt]
&\gt0
\end{align}
$$
This implies that the sequence
$$
b\_n=\sum\_{k=1}^n\frac1{\sqrt{k}}-2\sqrt{n+1}
$$
is increasing.

Note that
$$
a\_n-b\_n=\frac2{\sqrt{n\vphantom{+1}}+\sqrt{n+1}}
$$
which tends to $0$.

Thus any $b\_n$ is a lower bound for $a\_n$.