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Wolfram said that $$\sum_{k=1}^{\infty}\arctan\left(\frac{1}{k^2}\right)=\arctan\left(\frac{1-\cot\left(\frac{\pi}{\sqrt 2}\right)\tanh\left(\frac{\pi}{\sqrt 2}\right)}{1+\cot\left(\frac{\pi}{\sqrt 2}\right)\tanh\left(\frac{\pi}{\sqrt 2}\right)}\right).$$

I was wondering about a closed-form of $$S = \sum_{k=1}^{\infty}\arctan\left(\frac{1}{k^3}\right).$$

A numerical approximation of $S$ is $$S \approx 0.986791652613071125515794193830247643724471031136456434\dots$$

Is there a closed-form of $S$?

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  • $\begingroup$ $\Im\log\left[ \Gamma(i) \Gamma\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right) \Gamma\left(-\frac{\sqrt{3}}{2}-\frac{i}{2}\right) \right] -\frac{\pi}{2} $, WA will return a number matching yours. $\endgroup$ – achille hui Oct 11 '14 at 17:35
  • $\begingroup$ @achillehui Are you sure? For me it maches just for the first $26$ digits. How did you get this? $\endgroup$ – user153012 Oct 11 '14 at 17:40
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    $\begingroup$ I see, then your numerical approximation is probably not good enough. The key is $\tan^{-1}(i/k^3) = \Im\log(1 + i/k^3)$ and you use the infinite product expansion of gamma function $1/\Gamma(z) = z e^{\gamma z}\prod_{k=1}^\infty ( 1 + z/k) e^{-z/k}$ to reexpress you sum of logs in terms of product of 3 gamma function. $\endgroup$ – achille hui Oct 11 '14 at 17:48
  • $\begingroup$ @achillehui I've made the approximation with Mathematica. Now I updated the question and the evaluation by Maple seems correct. Could you write a more detailed proof as an answer? It would be great. $\endgroup$ – user153012 Oct 11 '14 at 22:09
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Notice for any positive number $x$, we have

$$\tan^{-1}(x) = \frac{1}{2i}\log\left(\frac{1+ix}{1-ix}\right) = \Im\log(1 + ix)$$

We can rewrite the sum at hand as

$$\sum_{k=1}^\infty \tan^{-1}\frac{1}{k^3} = \Im\left[ \sum_{k=1}^\infty \log\left(1 + \frac{i}{k^3}\right) \right] $$

For each $k$, we have the factorization

$$1 + \frac{i}{k^3} = 1 - \left(\frac{i}{k}\right)^3 = \prod_{j=0}^2 \left(1 - \frac{i\omega^j}{k}\right)$$ where $\omega = e^{i2\pi/3}$ is a primitive cubic root of unity.

Recall the infinite product expansion of gamma function

$$\frac{1}{\Gamma(z)} = z e^{\gamma z}\prod_{k=1}^\infty \left(1 + \frac{z}{k}\right) e^{-\frac{z}{k}} \quad\implies\quad \prod_{k=1}^\infty \left(1+\frac{z}{k}\right)e^{-\frac{z}{k}} = \frac{e^{-\gamma z}}{\Gamma(1+z)} $$

If we replace $z$ by $-i \omega^j$ for $j = 0,1,2$ and taking the product, we get

$$\begin{align} \prod_{k=1}^\infty\left(1 + \frac{i}{k^3}\right) &= \prod_{k=1}^\infty\prod_{j=0}^2\left[\left(1 - \frac{i\omega^j}{k}\right)e^{i\omega^j/k} \right] = \prod_{j=0}^2\prod_{k=1}^\infty\left[\left(1 - \frac{i\omega^j}{k}\right)e^{i\omega^j/k} \right]\\ & = \frac{e^{-\gamma( -i\sum\limits_{j=0}^2 \omega^j )}}{\prod\limits_{j=0}^2\Gamma(1 - i\omega^j)} = \frac{1}{ \Gamma(1-i) \Gamma\left(1 + \frac{\sqrt{3}}{2} + \frac{i}{2}\right) \Gamma\left(1 - \frac{\sqrt{3}}{2} + \frac{i}{2}\right)} \\ \end{align} $$ Taking logarithm on both sides, we will get something close to what we want.

The catch is the $\log$ function is not holomorphic over the whole complex plane. In general, the sum of logarithms is equal to the log of the product only up to some multiples of $2\pi$. i.e.

$$ \sum_{k=1}^\infty \log\left( 1 + \frac{i}{k^3}\right) = \log\left[ \prod_{k=1}^\infty \left( 1 + \frac{i}{k^3}\right) \right] + 2N\pi$$

for some unknown integer $N$.

Instead of determining what $N$ is, we will solve this problem in a different manner.

We use the fact in the sum of the log, the imaginary part for the $k \ge 2$ terms are small enough.
If we remove the $k = 1$ term from the sum, the sum of log will be equal to the log of product.
At the end, we have

$$\begin{align} \sum_{k=1}^\infty \tan^{-1}\frac{1}{k^3} &= \frac{\pi}{4} + \sum_{k=2}^\infty \tan^{-1}\frac{1}{k^3} = \frac{\pi}{4} + \Im\left[\sum_{k=2}^\infty\log\left(1 + \frac{i}{k^3}\right)\right]\\ &= \frac{\pi}{4} + \Im\log\left[\prod_{k=2}^\infty\left(1 + \frac{i}{k^3}\right)\right]\\ &= \frac{\pi}{4} - \Im\log\left[(1+i) \Gamma(1-i) \Gamma\left(1 + \frac{\sqrt{3}}{2} + \frac{i}{2}\right) \Gamma\left(1 - \frac{\sqrt{3}}{2} + \frac{i}{2}\right) \right]\\ &= \frac{\pi}{4} - \Im\log\left[(-1+i) \Gamma(-i) \Gamma\left( \frac{\sqrt{3}}{2} + \frac{i}{2}\right) \Gamma\left( - \frac{\sqrt{3}}{2} + \frac{i}{2}\right) \right]\\ \end{align}$$ According to WA, this is approximately $$0.986791652613071125515794193830247643724471031136456434028974\ldots$$ as expected.

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  • $\begingroup$ Thank you, really nice answer. Am I right, that with this method we could generalize the solution to $\sum_{k=1}^{\infty}\arctan(1/k^n)$ for some $n>1$ integer? $\endgroup$ – user153012 Oct 11 '14 at 23:44
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    $\begingroup$ @user153012 Yup, the solution can be generalized to all $n > 1$. $\endgroup$ – achille hui Oct 11 '14 at 23:59

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