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.

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$$
S_{n}\equiv\sum_{i = 1}^{n}\cosh\pars{1 \over \root{n + i}} - n
=
\sum_{i = n + 1}^{2n}\cosh\pars{1 \over \root{i}} - n= {\cal S}_{2n} - {\cal S}_{n} - n$$
$$
{\cal S}_{n} \equiv \sum_{i = 1}^{n}\cosh\pars{1 \over \root{i}}
$$

\begin{align}
{\cal S}_{n}&=
\sum_{i = 1}^{n}\bracks{1 + {1 \over 2i}}
+
\sum_{i = 1}^{n}\bracks{%
\cosh\pars{1 \over \root{i}} - 1 - {1 \over 2i}}
\\[3mm]&=
n + {1 \over 2}\,\ln\pars{n}
+
{1 \over 2}\
\overbrace{\bracks{\sum_{i = 1}^{n}{1 \over i} - \ln\pars{n}}}
^{\ds{\to \gamma\ \mbox{( Euler constant )} \atop \mbox{when}\ n \to \infty}}
+
\overbrace{\sum_{i = 1}^{n}\bracks{%
\cosh\pars{1 \over \root{i}} - 1 - {1 \over 2i}}}
^{\ds{\mbox{converges when}\ n \to \infty}}
\\
{\cal S}_{n} & \sim n + {1 \over 2}\,\ln\pars{n} + \mbox{constant}
\quad\mbox{when}\quad n \gg 1 
\end{align}

Then,
$$
S_{n} \sim \bracks{2n + {1 \over 2}\,\ln\pars{2n} + \mbox{constant}}
-
\bracks{n + \half\ln\pars{n} + \mbox{constant}} - n = \half\ln\pars{2}\quad\mbox{when}\quad n \gg 1
$$

$$\color{#0000ff}{\large%
\lim_{n \to \infty}S_{n} = \half\ln\pars{2}}
$$

