Find the limit of $S_n=\sum_{i=1}^n \big\{ \cosh\big(\!\!\frac{1}{\sqrt{n+i}}\!\big) -n\big\}$, as $n\to\infty$?

Question

$S_n=\sum_{i=1}^n\big\{ \cosh\big(\frac{1}{\sqrt{n+i}}\!\big) -n\big\}$ 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.

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ S_{n}\equiv\sum_{i = 1}^{n}\cosh\pars{1 \over \root{n + i}} - n = \sum_{i = n + 1}^{2n}\cosh\pars{1 \over \root{i}} - n= {\cal S}_{2n} - {\cal S}_{n} - n$$ $$ {\cal S}_{n} \equiv \sum_{i = 1}^{n}\cosh\pars{1 \over \root{i}} $$

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

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

Answer 2

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 $$ \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n} =\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\cdots+\frac{1}{1+\frac{n}{n}}\right) \to \int_0^1\frac{dx}{1+x}. $$

Answer 3

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 $$\lim_{n\to\infty}S_n=\frac 1 2\int_0^1\frac{dx}{1+x}=\frac{\log 2}2$$