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 $S$?

  • $\begingroup$ $\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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 3
    $\begingroup$ @user153012 Yup, the solution can be generalized to all $n > 1$. $\endgroup$ – achille hui Oct 11 '14 at 23:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.