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

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!

• Yes (+1) Alternating series, but is it convergent???? What happens if n goes to infinity? Commented Jan 14, 2014 at 16:56
• No, you are not allowed to get rid of those braces in this case! The conversion done by @CameronWilliams is wrong Commented Jan 14, 2014 at 17:02
• @CameronWilliams It has nothing to do with absolute convergence. In general you are allowed to set braces whereever you want, but not to remove them. Second thing is allowed only if the resulting list is convergent as well, which is obviously not the case here. Commented Jan 14, 2014 at 17:18
• It looks like the sum is $\tan^{-1}\left(\tanh\left(\frac{\pi}{2}\right)\right)$. Commented Jan 14, 2014 at 17:24
• This might be relevant (in the sense of "method" rather than "full solution") since we have the identity $\arctan (a)-\arctan(b)=\arctan(\frac{a-b}{1-ab})$ Commented Jan 14, 2014 at 17:42

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.

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

\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}

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}