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It is known that $$\sum_{n=1}^{\infty} \arctan \left(\frac{1}{n^{2}} \right) = \frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right). $$

Can we also find a closed form for the value of $$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \left(\frac{1}{n} \right)? $$

Unlike the other infinite series, this infinite series only converges conditionally.

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  • $\begingroup$ If you take the algorithmic for of arctan you have an infinite product but I don't think it helps. $\endgroup$ Commented Apr 30, 2014 at 20:49
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    $\begingroup$ why not answer your question yourself or leave it for others to answer. $\endgroup$
    – S L
    Commented Apr 30, 2014 at 20:57
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    $\begingroup$ You shouldn't delete the question; leave it up, so people who have the same question will be able to get an answer. $\endgroup$
    – Lost
    Commented Apr 30, 2014 at 21:30
  • $\begingroup$ Note the same series is mentioned for the sequence oeis.org/A265011 as it result in OEIS, page also provides close forms and alternate expression, however does not include any "how to ..." obviously. $\endgroup$
    – Machinato
    Commented Jun 11, 2016 at 20:15

6 Answers 6

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I have found a closed form expression for the series but it is sort of ugly, it involves gamma functions evaluated at complex arguments.

Regrouping the series into units of two, we have

$$\sum_{n=1}^\infty (-1)^{n-1}\tan^{-1}\frac{1}{n} = \sum_{k=1}^\infty a_k \quad\text{ where }\quad a_k = \tan^{-1}\frac{1}{2k-1} - \tan^{-1}\frac{1}{2k}.$$ Notice $\tan^{-1}(x) = \Im\log(1+i x)$ for real $x$, we can rewrite $a_k$ as

$$a_k = \Im\left\{\log\frac{1+\frac{i}{2k-1}}{1+\frac{i}{2k}}\right\} = \Im\left\{\log\frac{1+\frac{-1+i}{2k}}{1+\frac{i}{2k}}\right\} = \Im\left\{\log\frac{\left(1+\frac{-1+i}{2k}\right)e^{-\frac{-1+i}{2k}}}{\left(1+\frac{i}{2k}\right)e^{-\frac{i}{2k}}}\right\} $$ This implies up to some integer multiples of $2\pi$, we have

$$ \sum_{k=1}^\infty a_k = \Im\left\{ \log\frac{ e^{\gamma\frac{-1+i}{2}} \prod_{k=1}^\infty \left(1+\frac{-1+i}{2k}\right)e^{-\frac{-1+i}{2k}} }{ e^{\gamma\frac{i}{2}}\prod_{k=1}^\infty \left(1+\frac{i}{2k}\right)e^{-\frac{i}{2k}} } \right\} + 2N\pi $$ Using the infinite product expansion of Gamma function $$\frac{1}{\Gamma(z)} = z e^{\gamma z}\prod_{k=1}^\infty \left(1 + \frac{z}{k}\right) e^{-\frac{z}{k}} $$ and notice the $a_k$ are so small which forces $\displaystyle \left|\sum_{k=1}^\infty a_k\right| < 1$, we find the corresponding $N = 0$ and arrived at following closed form expression of the series.

$$\sum_{n=1}^\infty (-1)^{n-1}\tan^{-1}\frac{1}{n} = \Im\left\{\log\Gamma\left(1+\frac{i}{2}\right) - \log\Gamma\left(\frac12+\frac{i}{2}\right) \right\}\\ \approx 0.506670903216622981985255804783581512472843547347020582920002... $$

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  • $\begingroup$ Same answer, different methods and different representations. I think that's pretty nice. (+1) $\endgroup$
    – robjohn
    Commented Apr 30, 2014 at 23:48
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In the same spirit as this answer, note that $$ \log\left(\frac{n+i}n\right)=\frac12\log\left(1+\frac1{n^2}\right)+i\arctan\left(\frac1n\right) $$ Furthermore, using Gautschi's Inequality $$ \begin{align} \prod_{k=1}^{n-1}\frac{k+x}{k} &=\frac{\Gamma(n+x)}{\Gamma(1+x)\Gamma(n)}\\ &\sim\frac{n^x}{\Gamma(1+x)} \end{align} $$ Therefore, we get $$ \begin{align} \sum_{k=1}^{2n}(-1)^{k-1}\arctan\left(\frac1k\right) &=\mathrm{Im}\left(\log\left(\frac{1+i}{1}\frac{2}{2+i}\frac{3+i}{3}\frac{4}{4+i}\cdots\frac{2n}{2n+i}\right)\right)\\ &=\mathrm{Im}\left(\log\left(\frac{1+i}{1}\frac{2+i}{2}\frac{3+i}{3}\frac{4+i}{4}\cdots\frac{2n+i}{2n}\right)\right)\\ &-2\,\mathrm{Im}\left(\log\left(\frac{2+i}{2}\frac{4+i}{4}\cdots\frac{2n+i}{2n}\right)\right)\\ &=\mathrm{Im}\left(\log\left(\frac{1+i}{1}\frac{2+i}{2}\frac{3+i}{3}\frac{4+i}{4}\cdots\frac{2n+i}{2n}\right)\right)\\ &-2\,\mathrm{Im}\left(\log\left(\frac{1+\frac i2}{1}\frac{2+\frac i2}{2}\cdots\frac{n+\frac i2}{n}\right)\right)\\ &\sim\mathrm{Im}\left(\log\left(\frac{(2n)^i}{\Gamma(1+i)}\right)-2\log\left(\frac{n^{i/2}}{\Gamma(1+\frac i2)}\right)\right)\\ &=\log(2)+\mathrm{Im}\left(\log\left(\frac{\Gamma(1+\frac i2)^2}{\Gamma(1+i)}\right)\right) \end{align} $$ Therefore, $$ \begin{align} \sum_{k=1}^{2n}(-1)^{k-1}\arctan\left(\frac1k\right) &=\log(2)+\mathrm{Im}\left(\log\left(\frac{\Gamma(1+\frac i2)^2}{\Gamma(1+i)}\right)\right)\\[6pt] &=\log(2)-\mathrm{Im}\left(\log\binom{i}{i/2}\right)\\[9pt] &=\log(2)-\arg\binom{i}{i/2}\\[12pt] &\doteq0.506670903216622981985255804784 \end{align} $$

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    $\begingroup$ I would like to know what book you studied when you were still on school.... lol (+1) $\endgroup$
    – S L
    Commented Apr 30, 2014 at 23:51
  • $\begingroup$ I find it slightly amusing that Mathematica understands my second form. This gives the numerical answer I posted: N[Log[2]-Im[Log[Binomial[I,I/2]]],30] $\endgroup$
    – robjohn
    Commented May 1, 2014 at 0:12
  • $\begingroup$ Mathematica also likes N[Log[2]-Arg[Binomial[I,I/2]],30] $\endgroup$
    – robjohn
    Commented May 1, 2014 at 8:25
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Let $ \displaystyle S(a) = \sum_{n=1}^{\infty}(-1)^{n-1} \arctan \left(\frac{a}{n} \right)$.

Since $ \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n}{n^{2}+a^{2}}$ converges uniformly on $\mathbb{R}$,

$$ \begin{align} S'(a) &= \sum_{k=1}^{\infty} (-1)^{n-1} \frac{n}{a^{2}+n^{2}} \\ &= \frac{1}{2} \sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{1}{n-ia} + \frac{1}{n+ia} \right) \\ &= \frac{1}{2} \sum_{n=0}^{\infty} (-1)^{n} \left(\frac{1}{n+1-ia}+ \frac{1}{n+1+ia} \right). \end{align}$$

Then using the fact $$ \sum_{k=0}^{\infty} \frac{(-1)^{k}}{z+k} = \frac{1}{2} \Big[\psi\left( \frac{z+1}{2}\right) - \psi \left(\frac{z}{2} \right) \Big] $$ (where $\psi(z)$ is the digamma function), we have

$$ S'(a) = \frac{1}{4} \left[\psi \left(1- \frac{ia}{2} \right) - \psi\left(\frac{1}{2}- \frac{ia}{2} \right) + \psi\left(1+ \frac{ia}{2} \right) -\psi\left(\frac{1}{2}- \frac{ia}{2} \right)\right].$$

Integrating back we find

$$ S(a) = \frac{i}{2} \left[\log \Gamma \left(1-\frac{ia}{2} \right) - \log \Gamma \left(\frac{1}{2}-\frac{ia}{2} \right) - \log \Gamma \left(1+\frac{ia}{2} \right) + \log \Gamma \left(\frac{1}{2}+\frac{ia}{2} \right)\right] + C.$$

But since $S(0) = 0$, the constant of integration is $0$.

Therefore,

$$ \begin{align} S(1) &= \sum_{n=1}^{\infty}(-1)^{n-1} \arctan \left(\frac{1}{n} \right) \\ &= \frac{i}{2} \left[\log \Gamma \left(1-\frac{i}{2} \right) - \log \Gamma \left(\frac{1}{2}-\frac{i}{2} \right) - \log \Gamma \left(1+\frac{i}{2} \right) + \log \Gamma \left(\frac{1}{2}+\frac{i}{2} \right)\right] \\ &\approx 0.5066709032. \end{align}$$

The answer can be put in the same form as the answer given by achille hui by using the Schwarz reflection principle.

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  • $\begingroup$ Nice argument. I have used the $\psi$ function a number of times. As you say, the answer is essentially the same form as achille hui's, but the argument is unique. (+1) $\endgroup$
    – robjohn
    Commented May 1, 2014 at 1:55
  • $\begingroup$ @robjohn Thanks. $\endgroup$ Commented May 1, 2014 at 2:01
  • $\begingroup$ Compare with this answer that I gave in chat a while ago. You may need to install render MathJax to read it in the transcript. $\endgroup$
    – robjohn
    Commented May 1, 2014 at 8:39
  • $\begingroup$ @robjohn Is chat mainly for socializing, or are a lot of math questions asked and answered on there? $\endgroup$ Commented May 1, 2014 at 15:43
  • $\begingroup$ There is some socializing, but a lot of math that gets discussed there. $\endgroup$
    – robjohn
    Commented May 1, 2014 at 16:32
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\arctan\pars{1 \over n}:\ {\large ?}}$

$\ds{\Gamma\pars{z}}$ and $\ds{\Psi\pars{z}}$ are the Gamma and Digamma Functions, respectively. We'll use well known properties of them.

\begin{align} &\color{#00f}{\large\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\arctan\pars{1 \over n}} =\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\int_{0}^{1}{n\,\dd x \over x^{2} + n^{2}} \\[3mm]&=\int_{0}^{1}\bracks{\sum_{n = 1}^{\infty} \pars{-1}^{n + 1}\,{n \over n^{2} + x^{2}}}\,\dd x =\Re\int_{0}^{1}\bracks{\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n + x\ic}} \,\dd x \\[3mm]&=\Re\int_{0}^{1}\bracks{\sum_{n = 0}^{\infty}\pars{% {1 \over 2n + 1 + x\ic} - {1 \over 2n + 2 + x\ic}}}\,\dd x \\[3mm]&={1 \over 4}\Re\int_{0}^{1}\bracks{\sum_{n = 0}^{\infty} {1 \over \pars{n + 1/2 + x\ic/2}\pars{n + 1 + x\ic/2}}}\,\dd x \\[3mm]&={1 \over 4}\Re\int_{0}^{1} 2\bracks{\Psi\pars{1 + {x \over 2}\,\ic} - \Psi\pars{\half + {x \over 2}\,\ic}} \,\dd x \\[3mm]&=\half\Re\bracks{% -2\ic\ln\pars{\Gamma\pars{1 + {x \over 2}\,\ic}} + 2\ic\ln\pars{\Gamma\pars{\half + {x \over 2}\,\ic}}}_{0}^{1} \\[3mm]&=\Im\bracks{\ln\pars{\Gamma\pars{1 + \half\,\ic}} -\ln\pars{\Gamma\pars{\half + \half\,\ic}}} \\[3mm]&=\color{#00f}{\large% \Im\ln\pars{\Gamma\pars{1 + \ic/2} \over \Gamma\pars{1/2 + \ic/2}}} \approx 0.5067 \end{align}

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  • $\begingroup$ A more direct variant of Random Variable's answer (avoids differentiating and re-integrating). (+1) $\endgroup$
    – robjohn
    Commented May 1, 2014 at 1:53
  • $\begingroup$ @robjohn It's true. Any time I see $\large \arctan$ I remember the $\large 1/\left(x^{2} + 1\right)$-integral which always provides a nice sum. Thanks. $\endgroup$ Commented May 1, 2014 at 1:55
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This is just a possible way to proceed, and doesn't provide in any way a complete solution to the problem, thus I will put it as community wiki, and pray you to contribute if you have any insights.

The series converges (check!), inserting the Taylor series for $\arctan(x)$: $$\begin{array}{ll}\sum_{n=1}^\infty(-1)^{n+1}\arctan\left(\frac{1}{n}\right)&=\sum_{n=1}^\infty(-1)^{n+1}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\frac{1}{n^{2k+1}}\\&=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n^{2k+1}}\\&=\sum_{k=0}^\infty\frac{(-1)^k(1-2^{-2k})}{2k+1}\zeta(2k+1)\end{array}$$ where in the second equality we have used the fact that everything converges, and in the third we noticed that the second series in the second line is the Dirichlet eta function.

I don't know if this leads anywhere, but since the zeta function has been studied pretty heavily maybe there's some results somewhere that we can use to finish from here.

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This is not an answer, but a useful way to transform the expression and link it to another, more simple sum:

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

$$\arctan \frac{1}{2n}-\arctan \frac{1}{2n+1}=\arctan \frac{1}{4n^2+2n+1}$$

$$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \frac{1}{n}=\frac{\pi}{4}-\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2+2n+1}$$


From this question (ananswered by the way) we have an identity:

$$\sum_{n=0}^{N} \arctan \frac{1}{n^2+n+1}=\arctan(N+1)$$

Which means

$$\sum_{n=1}^{\infty} \arctan \frac{1}{n^2+n+1}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$$

Considering:

$$(2n-1)^2+2n-1+1=4n^2-2n+1$$

We get another identity (separating even and odd terms):

$$\sum_{n=1}^{\infty} \arctan \frac{1}{n^2+n+1}=\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2+2n+1}+\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2-2n+1}$$

And now we have:

$$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \frac{1}{n}=\sum_{n=1}^{\infty} \arctan \frac{1}{4n^2-2n+1}$$

The convergence of these two sums is slightly better than the original

enter image description here

If we take geometric mean of the last two expressions, it gives excellent convergence.

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