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.
 A: We have the approximation
$$
\cosh h = 1+\frac{h^2}{2!}+{\mathcal O}(h^4),
$$
which implies that
$$
\cosh \big((n+i)^{-1/2}\big)-1=\frac{1}{2n+2i}+{\mathcal O}\left(\frac{1}{n^2}\right).
$$
Thus
$$
\sum_{i=1}^n \cosh \big((n+i)^{-1/2}\big)-n=\frac{1}{2n+2}+\frac{1}{2n+4}+\cdots+\frac{1}{2n+2n}
+n\cdot{\mathcal O}\Big(\frac{1}{n^2}\Big)\to\frac{\ln 2}{2},
$$
as
$$
\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)=\ln 2.
$$
The last is due to the fact that
$$
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}
=\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdots+\frac{1}{1+\frac{n}{n}}\right) \to \int_0^1\frac{dx}{1+x}.
$$
A: 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
$$\lim_{n\to\infty}S_n=\frac 1 2\int_0^1\frac{dx}{1+x}=\frac{\log 2}2$$
A: $\newcommand{\+}{^{\dagger}}%
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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,
$$
S_{n} \sim \bracks{2n + {1 \over 2}\,\ln\pars{2n} + \mbox{constant}}
-
\bracks{n + \half\ln\pars{n} + \mbox{constant}} - n = \half\ln\pars{2}\quad\mbox{when}\quad n \gg 1
$$

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

