Prove that the limit of $a_n=\sum_{k=1}^n \sqrt\frac{1}{n^2+k^2}$ is $\ln(1+\sqrt{2})$ Let 
$$a_n=\sum_{k=1}^n \sqrt\frac{1}{n^2+k^2}$$Prove that the limit of the sequence  {$a_n$} is $\ln(1+\sqrt{2})$
 A: I'd start by factoring an $n^2$ out of the denominator, to get
$$
a_n=\sum_{k=1}^{n}\frac{1}{n}\sqrt{\frac{1}{1+\left(\frac{k}{n}\right)^2}}
$$
To me, this looks like we've taken $[0,1]$ and broken it down into $n$ pieces $[0,\tfrac{1}{n}]$, $(\tfrac{1}{n},\tfrac{2}{n}]$, etc. 
In this context, if we were setting up a Riemann sum for an integral, we would have $\Delta x=\frac{1}{n}$, and $x_i=\frac{k}{n}$ (where we commit the usual abuses of notation taught in integral calculus).  So, we could rewrite your sum as
$$
\sum_{k=1}^{n}\sqrt{\frac{1}{1+x_i^2}}\,\Delta x.
$$
Is it clear, in this form, what integral this Riemann sum would represent?
A: Let
$$f(x)=\frac{1}{\sqrt{1+x^2}}$$
then
$$a_n=\frac{1-0}{n}\sum_{k=1}^nf(0+\frac{k(1-0)}{n})$$
$f$ is continuous at $[0,1]$ thus
$$L=\lim_{n\to +\infty}a_n=\int_0^1f(x)dx$$
with the change $x=sh(t)$, we get
$L=argsh(1)$ and
$$\frac{(e^L-e^{-L})}{2}=1$$
which becomes
$$e^{2L}-2e^L-1=0.$$
thus
$e^L=1+\sqrt{2}$ and finally
$$\color{green}{L=Log(1+\sqrt{2}).}$$
