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: 
**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 $a\_n$.
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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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
>  $\ds{0 <
> {1 \over 2}\int\_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x <
> {1 \over 2}\int\_{n}^{\infty}{\dd x \over x^{3/2}} = {1 \over \root{n}}
> \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\LARGE\to}\,\,\,\color{red}{\large 0}}$
> 
> 
> 

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