Evaluate the sum of series of arctan and find limit

Question

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?

Answer 1

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

$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$

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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

$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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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}