TITLE: Find the value of : $\lim_{n \rightarrow \infty} \prod_{k=1}^n \left(1+\ln\left(\frac{k+\sqrt{k^2+n^2}}{n}\right)^{\frac{1}{n}}\right)$
QUESTION [5 upvotes]: Compute $\displaystyle \lim_{n \rightarrow \infty} \prod_{k=1}^n \left(1+\ln\left(\frac{k+\sqrt{k^2+n^2}}{n}\right)^{\frac{1}{n}}\right)$ 

Note that $\frac{k+\sqrt{k^2+n^2}}{n}\geq 1$ so we're dealing with positive terms.
Taking $\log$ of the product, we're interested in the limit of $\displaystyle \sum_{k=1}^n \ln\left(  1+ \exp \left(\frac 1n\ln\ln \left( \frac kn + \sqrt{\left( \frac{k}{n} \right) ^2 +1}\right)\right) \right)$
which look very much like a Riemann sum.
Setting $f(x) =\ln\ln(x+\sqrt{x^2+1})$, the sum rewrite as $\displaystyle \sum_{k=1}^n \ln\left(  1+ \exp \left(\frac 1nf \left(\frac kn\right)\right) \right)$
I've been trying to use the usual bounds on $\ln$ and $\exp$ to exhibit a Riemann sum, but it's quite messy.

REPLY [3 votes]: Call $$a(n,k)=\ln \left( \frac{k}{n}+\sqrt{\frac{k^2}{n^2}+1 } \right)$$
and notice that $|a(n,k)|\le C$ for some fixed constant $C$.
We want to evaulate
$$S_n=\sum_{k=1}^n \ln(1+1/n\ln(\frac{1}{n}a(n,k))).$$
Using $\log(1+x)=x+O(x^2)$, we can write
$$S_n=\sum_{k=1}^n \frac{1}{n}a(n,k)+O(\frac{1}{n}a(n,k))=\sum_{k=1}^n \frac{1}{n}a(n,k)+O(\frac{1}{n^2})=R_n+O(1/n)$$
where $R_n$ is the n-th Riemann sum of the integral
$$\int_0^1 \ln(x+\sqrt{x^2+1}) dx=I.$$
Hence 
$$\lim_{n \to \infty} S_n=I$$
and then the original product is $e^I$.