Proving sequence is bounded below? I have the function: 
$$
a_{n}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots +\frac{1}{\sqrt{n}}-2\sqrt{n}
$$

How could I go about proving that this sequence is bounded from below? When plotting it, I can see it converges to around -1.45, but I can't figure out how to try and prove it's bounded arithmetically. I was trying to figure out how to manipulate it to use the arithmetic/geometric mean inequality, but had no such luck.

Thanks!
hint: $\dfrac{1}{\sqrt{k}}= \dfrac{2}{2\sqrt{k}} \ge \dfrac{2}{\sqrt{k+1}+\sqrt{k}}= 2\sqrt{k+1}- 2\sqrt{k}$, and take the sum for $k$ runs from $1$ to $n-1$.
If
$a_{n}=\sum_{k=1}^n \frac1{\sqrt{k}}-2\sqrt{n}
$,
then

$\begin{array}\\
d(n)
&amp;=a_{n+1}-a_n\\
&amp;=\dfrac1{\sqrt{n+1}}-2(\sqrt{n+1}-\sqrt{n})\\
&amp;=\dfrac1{\sqrt{n+1}}-2(\sqrt{n+1}-\sqrt{n})\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\
&amp;=\dfrac1{\sqrt{n+1}}-2\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\\
&amp;&lt;\dfrac1{\sqrt{n+1}}-2\dfrac{1}{2\sqrt{n+1}}\\
&amp;=0\\
\text{and}\\
d(n)
&amp;&gt;\dfrac1{\sqrt{n+1}}-2\dfrac{1}{2\sqrt{n}}\\
&amp;=\dfrac1{\sqrt{n+1}}-\dfrac{1}{\sqrt{n}}\\
&amp;=\dfrac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}\sqrt{n+1}}\\
&amp;=\dfrac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}\sqrt{n+1}}\dfrac{\sqrt{n}+\sqrt{n+1}}{\sqrt{n}+\sqrt{n+1}}\\
&amp;=\dfrac{-1}{\sqrt{n}\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})}\\
&amp;&gt;\dfrac{-1}{2n\sqrt{n}}\\
\text{so}\\
a_{n+k}-a_n
&amp;=\sum_{j=1}^k (a(n+j)-a(n+j-1))\\
&amp;&lt; 0\\
\text{and}\\
a_{n+k}-a_n
&amp;=\sum_{j=1}^k (a(n+j)-a(n+j-1))\\
&amp;&gt;\sum_{j=1}^k \dfrac{-1}{2(n+j-1)\sqrt{n+j-1}}\\
\end{array}
$

Therefore
$a_n$ is a decreasing sequence
which is bounded below
since
$\sum \dfrac1{n^{3/2}}
$
converges.

For an explicit bound,
since
$\dfrac1{x^{3/2}}
$
is decreasing for
$x \ge 1$,
$\dfrac1{x^{3/2}}
\lt \int_{x-1}^x \dfrac{dt}{t^{3/2}}
$
so

$\begin{array}\\
\sum_{j=1}^k \dfrac{1}{(n+j-1)\sqrt{n+j-1}}
&amp;\lt\sum_{j=1}^k \int_{n+j-2}^{n+j-1} \dfrac{dt}{t^{3/2}}\\
&amp;\lt\int_{n-1}^{n+k-1} \dfrac{dt}{t^{3/2}}\\
&amp;\lt\int_{n-1}^{\infty} \dfrac{dt}{t^{3/2}}\\
&amp;\lt\dfrac{2}{\sqrt{n-1}}\\
\end{array}
$

so

$\begin{array}\\
a_{n+k}-a_n
&amp;&gt;\sum_{j=1}^k \dfrac{-1}{2(n+j-1)\sqrt{n+j-1}}\\
&amp;\gt\dfrac{-1}{\sqrt{n-1}}\\
\text{or}\\
a_{n+k}
&amp;&gt;a_n-\dfrac{1}{\sqrt{n-1}}\\
\end{array}
$

As is often the case,
nothing here is original
and I worked it out
as I went along.