# A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$

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.

• If you take the algorithmic for of arctan you have an infinite product but I don't think it helps. – Ali Caglayan Apr 30 '14 at 20:49
• You shouldn't delete the question; leave it up, so people who have the same question will be able to get an answer. – Lost Apr 30 '14 at 21:30
• 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. – Machinato Jun 11 '16 at 20:15

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.

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

• Same answer, different methods and different representations. I think that's pretty nice. (+1) – robjohn Apr 30 '14 at 23:48

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}

• I would like to know what book you studied when you were still on school.... lol (+1) – Santosh Linkha Apr 30 '14 at 23:51
• 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] – robjohn May 1 '14 at 0:12
• Mathematica also likes N[Log[2]-Arg[Binomial[I,I/2]],30] – robjohn May 1 '14 at 8:25

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.

• 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) – robjohn May 1 '14 at 1:55
• @robjohn Thanks. – Random Variable May 1 '14 at 2:01
• 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. – robjohn May 1 '14 at 8:39
• @robjohn Is chat mainly for socializing, or are a lot of math questions asked and answered on there? – Random Variable May 1 '14 at 15:43
• There is some socializing, but a lot of math that gets discussed there. – robjohn May 1 '14 at 16:32

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

$\ds{\Gamma\pars{z}}$ 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}

• A more direct variant of Random Variable's answer (avoids differentiating and re-integrating). (+1) – robjohn May 1 '14 at 1:53
• @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. – Felix Marin May 1 '14 at 1:55

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.

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

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