Proof the series is finite using following inequality Let 
$$a_n=\frac1{\sqrt1}+\frac1{\sqrt 2}+\ldots +\frac1{\sqrt n}-2\sqrt n $$
For the task to prove that $$\tag1-2\le a_n\le -1 $$
I was given the hint
$$\tag2\sqrt{k+1}-\sqrt k<\frac1{2\sqrt k},\qquad \forall k\in\mathbb N^* $$
I managed to prove $(1)$ by other methods but woonder how to proof it actually using $(2)$.
 A: First, let us prove (2), that is 
\begin{equation*}
\sqrt{k+1}-\sqrt{k}\leq \frac{1}{2\sqrt{k}},\ for\ k\geq 1.
\end{equation*}
\begin{eqnarray*}
\sqrt{k+1}-\sqrt{k} &=&\frac{\left( \sqrt{k+1}-\sqrt{k}\right) \left( \sqrt{%
k+1}+\sqrt{k}\right) }{\left( \sqrt{k+1}+\sqrt{k}\right) } \\
&=&\frac{1}{\left( \sqrt{k+1}+\sqrt{k}\right) } \\
&\leq &\frac{1}{\left( \sqrt{k}+\sqrt{k}\right) } \\
&=&\frac{1}{2\sqrt{k}}.
\end{eqnarray*}
Now let us prove the inequality
\begin{equation*}
-2\leq a_{n}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}%
-2\sqrt{n}.
\end{equation*}
Consider the inequality already proved, for $k=1,$ 2, $\cdots ,(n-1),$ and
write then vertically
\begin{eqnarray*}
\sqrt{2}-\sqrt{1} &\leq &\frac{1}{2\sqrt{1}} \\
\sqrt{3}-\sqrt{2} &\leq &\frac{1}{2\sqrt{2}} \\
&&\vdots  \\
\sqrt{n}-\sqrt{n-1} &\leq &\frac{1}{2\sqrt{n-1}}
\end{eqnarray*}
Now add them all, they telescope (!), that is $\sqrt{2}$ of the first line
cancel with $\sqrt{2}$ of the second line, and so on, until $\sqrt{n-1}$ of
the (n-2)$^{th}$ line which cancel with $\sqrt{n-1}$ of the last line, to get%
\begin{equation*}
\sqrt{n}-\sqrt{1}\leq \frac{1}{2\sqrt{1}}+\frac{1}{2\sqrt{2}}+\cdots +\frac{1%
}{2\sqrt{n-1}}
\end{equation*}
then
\begin{equation*}
\sqrt{n}-1\leq \frac{1}{2}\left( \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}%
+\cdots +\frac{1}{\sqrt{n-1}}\right) 
\end{equation*}
then
\begin{equation*}
2\sqrt{n}-2\leq \left( \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1%
}{\sqrt{n-1}}\right) 
\end{equation*}
which implies
\begin{equation*}
-2\leq \left( \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{%
n-1}}\right) -2\sqrt{n}=a_{n}.
\end{equation*}
As was to be shown.
It remains to prove the right inequality.
${\bf UPDATE}$ To prove that 
\begin{equation*}
a_{n}\leq -1,\ for\ all\ n\geq 1.
\end{equation*}
I was not able to prove it using the inequality 
\begin{equation*}
\sqrt{k+1}-\sqrt{k}\leq \frac{1}{2\sqrt{k}},\ for\ k\geq 1.
\end{equation*}
However, if we compute 
\begin{equation*}
a_{n+1}-a_{n}=\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n+1}\left( \sqrt{n+1}+\sqrt{n}%
\right) }<0.
\end{equation*}
So the sequence (a$_{n})$ is decreasing, and since $a_{1}=\frac{1}{\sqrt{1}}%
-2\sqrt{1}=-1$ then
\begin{equation*}
a_{n}\leq a_{1}=-1,\ \ for\ all\ n\geq 1.
\end{equation*}
A: $a_{n+1} = \displaystyle \sum_{k=1}^{n+1}\dfrac{1}{\sqrt{k}}-2\sqrt{n+1} = a_n + \dfrac{1}{\sqrt{n+1}}- 2\left(\sqrt{n+1}- \sqrt{n}\right) \leq a_n \leq -1$. That is how you use it via induction.
A: Using the hint we obtain a telescoping sum
$$\begin{align}a_n&>2(\sqrt2-\sqrt1)+2(\sqrt3-\sqrt2)+\ldots +2(\sqrt{n+1}-\sqrt n)\,-\,2\sqrt n \\&=2(\sqrt {n+1}-\sqrt n-1)\\&>-2\end{align}$$
