Find the limit of $S_n=\sum_{i=1}^n \big\{ \cosh\big(\!\!\frac{1}{\sqrt{n+i}}\!\big) -n\big\}$, as $n\to\infty$? $S_n=\sum_{i=1}^n\big\{ \cosh\big(\frac{1}{\sqrt{n+i}}\!\big) -n\big\}$ as $n\to\infty$
I stumbled on this question as an reading about Riemannian sums as in 
$$
\int_a^b f(x)\,dx =\lim_{x\to \infty}\frac{1}{n}\sum_{k=1}^n\, f\Big(a+k\frac{b-a}{n}\Big).
$$ 
With numerical approximations, it seems as though $$\lambda=\lim_{x\to \infty}S_n =\log(\sqrt2).$$ But can we apply the Riemannian definition to this sum (by fetching a suitable function)? On the other hand, I tried the squeeze theorem on the sum to yield $0.5<\lambda<1$ , but still no closed form.

We have
$$S_n=\sum_{i=1}^n \cosh\left(\frac{1}{\sqrt{n+i}}\right) -n=\sum_{i=1}^n \left(\cosh\left(\frac{1}{\sqrt{n+i}}\right) -1\right)$$ 
and by the Taylor-Lagrange inequality we have
$$\left|\cosh\left(\frac{1}{\sqrt{n+i}}\right) -1-\frac{1}{2(n+i)}\right|\le\frac{C}{n^{3/2}}$$
hence
$$\left|S_n-\sum_{i=1}^n\frac{1}{2(n+i)}\right|\le \frac{C}{n^{1/2}}\to0$$
so by Riemann series we have
$$\lim_{n\to\infty}S_n=\frac 1 2\int_0^1\frac{dx}{1+x}=\frac{\log 2}2$$
