Evaluate the sum of series of arctan and find limit I try to find the value of following limit:
$$
\lim _{n \rightarrow \infty} 2^{-n/2} \sum_{j=2^{n}+1}^{2^{n+1}} \tan ^{-1}\left(\frac{1}{\sqrt{j-1}}\right)
$$
But I cannot evaluate the summation. Can anybody help?
 A: METHODOLOGY $1$:  Simple Bounds and Using the Squeeze Theorem
Since the arctangent is monotonically increasing, $\arctan(1/\sqrt{x-1})$ is monotonically decreasing.  Hence we can assert that
$$\int_{N}^{M+1}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx\le \sum_{j=N}^M\arctan\left(\frac1{\sqrt{j-1}}\right)\le \int_{N-1}^{M}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx\tag1$$
The antiderivative of $\arctan\left(\frac1{\sqrt{x-1}}\right)$ can be written as
$$\int \arctan\left(\frac1{\sqrt{x-1}}\right)\,dx=\sqrt{x-1}+x\arctan\left(\frac1{\sqrt{x-1}}\right)+C\tag2$$
For $N=2^n+1$ and $M=2^{n+1}$ in $(2)$, we find that
$$\begin{align}
\int_{2^{n}+1}^{2^{n+1}+1} \arctan\left(\frac1{\sqrt{x-1}}\right)\,dx&=2^{n/2}(\sqrt 2-1)\\\\
&+(2^{n+1}+1)\arctan\left(\frac1{\sqrt{2}2^{n/2}}\right)\\\\
&-(2^n+1)\arctan\left(\frac1{2^{n/2}}\right)\tag3
\end{align}$$
Dividing $(3)$ by $2^{n/2}$ and letting $n\to\infty$ reveals that
$$\liminf_{n\to\infty}2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)\ge 2(\sqrt 2-1)\tag4$$
Similarly, evaluation of the right-hand side of $(1)$ reveals
$$\limsup_{n\to\infty}2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)\le 2(\sqrt 2-1)\tag5$$
Finally, putting $(4)$ and $(5)$ together yields the coveted limit
$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$


METHODOLOGY $2$:  Use on the Euler-Maclaurin Summation Formula
From the Euler-Maclaurin Summation formula we have
$$\begin{align}
2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)&=2^{-n/2}\int_{2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx+o(1)\\\\
&=2^{-n/2}\left(\sqrt{2^{n+1}-1}-\sqrt{2^n} \right)\\\\
&+2^{n/2+1}\arctan\left(\frac1{\sqrt{2^{n+1}-1}}\right)\\\\
&-(2^{n/2}+1)\arctan\left(\frac1{\sqrt{2^{n}}}\right)+o(1)\tag6
\end{align}$$
Letting $n\to \infty$ in $(6)$, we find that
$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$
A: Using generalized harmonic numbers.
Since $n$is large, $j$ is large and, by Taylor expansion
$$\tan ^{-1}\left(\frac{1}{\sqrt{j-1}}\right)=\frac{1}{\sqrt j}+O\left(\frac{1}{j^{3/2}}\right)$$
$$S_n=\sum_{j=2^{n}+1}^{2^{n+1}} \tan ^{-1}\left(\frac{1}{\sqrt{j-1}}\right)\sim \sum_{j=2^{n}+1}^{2^{n+1}} \frac{1}{\sqrt j}=H_{2^{n+1}}^{\left(\frac{1}{2}\right)}-H_{2^n}^{\left(\frac{1}{2}\right)} $$
Now, using
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+O\left({\frac{1}{p^{1/2}}}\right)$$
$$H_{2^{n+1}}^{\left(\frac{1}{2}\right)}-H_{2^n}^{\left(\frac{1}{2}\right)}\sim 2{\sqrt{2^{n+1}}}-{2}{\sqrt{2^{n}}}=\left(\sqrt{2}-1\right) 2^{\frac{n}{2}+1} $$ and then the result already given by @Mark Viola.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{2^{-n/2}
\sum_{\vphantom{\large A}j\ =\ 2^{n}\ +\ 1}^{2^{\,n + 1}} \arctan\pars{1 \over \root{j - 1}}}
\\[5mm] &\
\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,\,
2^{-n/2}
\sum_{\vphantom{\large A}j\ =\ 2^{n}\ +\ 1}^{2^{\,n + 1}} {1 \over \root{j}}
\\[5mm] = &\
2^{-n/2}\pars{%
\sum_{j\ =\ 1}^{2^{\,n + 1}} {1 \over \root{j}} -
\sum_{j\ =\ 1}^{2^{\,n}} {1 \over \root{j}}}
\\[5mm] = &\
2^{-n/2}\left\{%
\bracks{\zeta\pars{1 \over 2} + 2\root{2^{n + 1}} +
{1 \over 2}\int_{2^{n + 1}}^{\infty}{\braces{x} \over x^{3/2}}\dd x}
\right.
\\[2mm] &\ \phantom{2^{-n/2}}
-\left.\bracks{\zeta\pars{1 \over 2} + 2\root{2^{n}} +
{1 \over 2}\int_{2^{n}}^{\infty}{\braces{x} \over x^{3/2}}\dd x}\right\}
\end{align}
In the last expression I used a Zeta Function Identity.
Then,
\begin{align}
&\bbox[5px,#ffd]{2^{-n/2}
\sum_{\vphantom{\large A}j\ =\ 2^{n}\ +\ 1}^{2^{\,n + 1}} \arctan\pars{1 \over \root{j - 1}}}
\\[5mm] &\ \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,\,
2\root{2} - 2 -\
\underbrace{{1 \over 2^{n/2 + 1}}\int_{2^{n}}^{2^{n} + 1}{\braces{x} \over x^{3/2}}\dd x}
_{\ds{\to \color{red}{\large 0}\ \mbox{as}\ n\ \to\ \infty}}
\\[5mm] \implies &\
\bbox[5px,#ffd]{\lim_{n \to \infty}\,\,\bracks{2^{-n/2}
\sum_{\vphantom{\large A}j\ =\ 2^{n}\ +\ 1}^{2^{\,n + 1}} \arctan\pars{1 \over \root{j - 1}}}}
\\[2mm] = &\
\bbx{2\root{2} - 2} \approx 0.8284 \\ &
\end{align}

Note that
\begin{align}
& 0 < \verts{{1 \over 2^{n/2 + 1}}\int_{2^{n}}^{2^{n} + 1}{\braces{x} \over x^{3/2}}\dd x} <
{1 \over 2^{n/2 + 1}}\int_{2^{n}}^{2^{n} + 1}
{\dd x \over x^{3/2}}
\\[5mm] = &\
{2^{-n} - \pars{2^{n} + 1}^{-1/2} \over 2^{n/2}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,\color{red}{\large 0}
\end{align}
