Investigate convergence of the series Investigate convergence of the series:
$$\left( \frac{n^2+3n+10}{n^2+5n+17} \right)^{n^2 (\sqrt{n+1}-\sqrt{n-1})}$$
It should be solvable with simple manipulations with the formula, i guess, but how to do that?
 A: Let $a_n$ be the above term.  Note first that
$$\frac{n^2+3 n+10}{n^2+5 n+17} = \left (1+\frac{3}{n}+\frac{10}{n^2} \right ) \left (1+\frac{5}{n}+\frac{17}{n^2} \right )^{-1} = 1-\frac{2}{n}+\frac{3}{n^2}+O\left (\frac1{n^3}\right )$$
Also,
$$n^2 \left ( \sqrt{n+1}-\sqrt{n-1}\right ) = n^{3/2} \left (1+\frac{3}{8 n^2} + O\left (\frac1{n^3}\right )\right ) $$
Then
$$\begin{align}\log{a_n} &= n^2 \left ( \sqrt{n+1}-\sqrt{n-1}\right ) \log{ \left [  1-\frac{2}{n}+\frac{3}{n^2}+O\left (\frac1{n^3}\right )\right]} \\ &= n^{3/2} \left (1+\frac{3}{8 n^2} + O\left (\frac1{n^3}\right )\right ) \left (-\frac{2}{n} + \frac1{n^2} +  O\left (\frac1{n^3}\right ) \right )\\ &= -2 \sqrt{n} + \frac1{\sqrt{n}} + O \left ( n^{-3/2} \right ) \end{align}$$
which means that the general term in the series behaves as
$$e^{-2 \sqrt{n}} \left ( 1+ \frac1{\sqrt{n}} \right ) $$
for large $n$.  Thus the series converges.
A: Note that
$$
\sqrt{n+1}-\sqrt{n-1}=\frac2{\sqrt{n+1}+\sqrt{n-1}}\ge\frac1{\sqrt{n+1}}
$$
and by cross-multiplication, for $n\ge20$,
$$
\frac{n^2+3n+10}{n^2+5n+17}\le\frac{n}{n+2}
$$
Therefore, using Bernoulli's Inequality,
$$
\begin{align}
\left(\frac{n^2+3n+10}{n^2+5n+17}\right)^{n^2(\sqrt{n+1}-\sqrt{n-1})}
&\le\left(\frac1{1+\frac2n}\right)^{n^2/\sqrt{n+1}}\\
&\le\frac1{1+\frac{2n}{\sqrt{n+1}}}\\[4pt]
&\to0
\end{align}
$$

Upon reading the question again, I am not certain whether you were wanting to know whether the terms above converge or whether their sum converges. Using the inequality
$$
1+x\le e^x
$$
we get that
$$
\begin{align}
\left(\frac{n^2+3n+10}{n^2+5n+17}\right)^{n^2(\sqrt{n+1}-\sqrt{n-1})}
&\le\left(1-\frac2{n+2}\right)^{n^2/\sqrt{n+1}}\\
&\le e^{-\frac{2n^2}{(n+2)\sqrt{n+1}}}\\[9pt]
&\le e^{-2\sqrt{n-5}}
\end{align}
$$
since $\frac{2n^2}{(n+2)\sqrt{n+1}}\ge2\sqrt{n-5}$. Thus, this term decays faster than any power of $n$. Therefore, even the sum converges.
