<s>[INST] 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?
 [/INST] 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
$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$
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