Evaluate $\sum_{n=1}^{\infty} \left( \arctan \frac{4n - 1}{2} - \arctan \frac{4n - 3}{2} \right)$

Question

I got stuck on the following series: $$ \sum_{n=1}^{\infty} \left( \arctan \frac{4n - 1}{2} - \arctan \frac{4n - 3}{2} \right). $$ I can't seem to make an approach because there's $-3$ not $+3$. Please help!

Answer 1

Let $a_n$ be the $n^{th}$ term of the series at hand. We have $$\begin{align} a_n = & \tan^{-1}\frac{4n-1}{2} - \tan^{-1}\frac{4n-3}{2}\\ = & \tan^{-1}\left(\frac{\frac{4n-1}{2}-\frac{4n-3}{2}}{1 + \frac{4n-1}{2}\frac{4n-3}{2}}\right) = \tan^{-1}\left(\frac{1}{(2n-1)^2 + \frac34}\right)\\ \end{align}$$ Notice $\;\tan^{-1}(x) = \Im\log(1 + ix)\;$ for real $x$, we can rewrite $a_n$ as

$$a_n = \Im\left\{\log\left( 1 + \frac{i}{(2n-1)^2 + \frac34}\right)\right\} = \Im\left\{\log\left( 1 + \frac{\frac34 + i}{(2n-1)^2}\right)\right\} $$

Compare this with the factors in the infinite product expansion of $\cosh x$:

$$\cosh x = \prod_{n=1}^{\infty}\left(1 + \frac{4x^2}{(2n-1)^2\pi^2}\right)$$ We find$\color{blue}{^{[1]}}$

$$\begin{align} \sum_{n=1}^\infty a_n = &\Im\left\{\log\cosh\left(\frac{\pi}{2}\sqrt{\frac34+i}\right)\right\} = \Im\left\{\log\cosh\left[\frac{\pi}{2}\left(1 + \frac{i}{2}\right)\right]\right\}\\ = & \Im\left\{\log\left[\cosh\frac{\pi}{2} \cos\frac{\pi}{4} + \sinh\frac{\pi}{2}\sin\frac{\pi}{4}i\right]\right\} = \Im\left\{\log\left[ 1 + \tanh\frac{\pi}{2} i\right]\right\}\\ = & \tan^{-1}\left[\tanh\left(\frac{\pi}{2}\right)\right] \end{align}$$

Notes

• $\color{blue}{[1]}$ Given any two complex numbers $u$ and $v$, $\log(uv)$ need not equal to $\log u + \log u$ in general. Instead, we have $$\log(uv) = \log u + \log v + i2\pi N$$ for some integer $N$. So in principle, $$\begin{align} a_n &= \Im\left\{\log\left( 1 + \frac{\frac34 + i}{(2n-1)^2}\right)\right\}\\ \implies \sum_{n=1}^\infty a_n &= \Im\left\{\log\prod_{n=1}^\infty\left( 1 + \frac{\frac34 + i}{(2n-1)^2}\right)\right\} + 2\pi N \end{align}$$ for some integer $N$ only. However, $a_n$ is small enough and the sum falls within the range $(-\frac{\pi}{2},\frac{\pi}{2})$, the $N$ here is actually zero. The naive looking replacement: $$\sum_{n=1}^\infty a_n \quad\longrightarrow\quad \Im\left\{\log\cosh\left(\frac{\pi}{2}\sqrt{\frac34+i}\right)\right\}$$ does work.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}\bracks{% \arctan\pars{4n - 1 \over 2} - \arctan\pars{4n - 3 \over 2}}:\ {\large ?}}$

\begin{align}&\color{#c00000}{\sum_{n = 1}^{\infty}\bracks{% \arctan\pars{4n - 1 \over 2} - \arctan\pars{4n - 3 \over 2}}} =\sum_{n = 0}^{\infty} \int_{\pars{4n + 1}/2}^{\pars{4n + 3}/2}{\dd x \over x^{2} + 1} \end{align}

With $\ds{x \equiv {4n + 1 \over 2} + \xi}$: \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}\bracks{% \arctan\pars{4n - 1 \over 2} - \arctan\pars{4n - 3 \over 2}}} =\sum_{n = 0}^{\infty} \int_{0}^{1}{\dd\xi \over \bracks{\pars{4n + 1}/2 + \xi}^{2} + 1} \\[5mm] = &\ \int_{0}^{1}\sum_{n = 0}^{\infty} {1 \over \pars{2n + 1/2 + \xi + \ic}\pars{2n + 1/2 + \xi - \ic}}\,\dd\xi \\[5mm] = &\ {1 \over 4}\int_{0}^{1}\sum_{n = 0}^{\infty} {1 \over \pars{n + 1/4 + \xi/2 + \ic/2}\pars{n + 1/4 + \xi/2 - \ic/2}}\,\dd\xi \\[5mm] = &\ {1 \over 4}\int_{0}^{1} {\Psi\pars{1/4 + \xi/2 + \ic/2} - \Psi\pars{1/4 + \xi/2 - \ic/2} \over \ic}\,\dd\xi \\[5mm] = &\ \half\,\Im\int_{0}^{1}\Psi\pars{1/4 + \xi/2 + \ic/2}\,\dd\xi \\[5mm] = &\ \Im\ln\pars{\Gamma\pars{3/4 + \ic/2} \over \Gamma\pars{1/4 + \ic/2}} = \Im\ln\pars{% \Gamma\pars{{3 \over 4} + {\ic \over 2}}\Gamma\pars{{1 \over 4} - {\ic \over 2}}} \qquad\qquad\qquad\quad\pars{1} \\[5mm] = &\ \Im\ln\pars{\pi\root{2} \over \cosh\pars{\pi/2} - \ic\sinh\pars{\pi/2}} = \arctan\pars{\tanh\pars{\pi \over 2}} \qquad\qquad\qquad\qquad\quad\pars{2} \end{align}

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In $\pars{1}$ and $\pars{2}$ we used identity ${\bf\mbox{6.1.32}}$ of $\large\mbox{this table}$. \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty} \bracks{% \arctan\pars{4n - 1 \over 2} - \arctan\pars{4n - 3 \over 2}}} \\[5mm] = &\ \bbox[10px,border:1px groove navy]{\arctan\pars{\tanh\pars{\pi \over 2}}} \approx 0.7422 \end{align}