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Here is a problem that I ran into. I seriously doubt if there is a closed form but you never know.

Evaluate the series

$$\mathcal{S} = \sum_{n=1}^\infty \arctan \left ( \frac 1 {n^3+n^2+n+1} \right) $$

I searched in vain to attack it using telescopic summation but I failed miserably. Then I remembered the following technique. Since ${\rm Im} \log (1+ix) = \arctan x$ we can express the sum as follows

\begin{align*} \sum_{n=1}^\infty \arctan \left ( \frac{1}{n^3+n^2+n+1} \right ) &= \sum_{n=1}^\infty \arctan \left [ \frac{1}{(n+1)(n^2+1)} \right ] \\ &= \sum_{n=1}^\infty \operatorname{Im} \left [ \log \left ( 1 + \frac{i}{(n+1)(n^2+1)} \right ) \right ] \\ &= \operatorname{Im} \log \left [ \prod_{n=1}^\infty \left ( 1 + \frac{i}{(n+1)(n^2+1)} \right ) \right ] \end{align*}

I tried to combine it with the famous Euler product

$$ \frac{\sin \pi z}{\pi z} = \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) \tag{1} $$

but I see no connection. So, is there a possible way to evaluate it?

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  • $\begingroup$ Possibly partial fractions plus the identity $\arctan a + \arctan b = \arctan \dfrac{a+b}{1-ab}$ will do it. $\qquad$ $\endgroup$ Oct 29, 2017 at 19:29
  • $\begingroup$ I guess we need some creativity to make what you suggest work ... ! $\endgroup$
    – Tolaso
    Oct 29, 2017 at 19:36
  • $\begingroup$ Don't know if it can be useful $$\arctan \frac{1}{n^3+n^2+n+1}=\arctan\left(\frac{1}{2} \left(1-\sqrt{4 n^3 +4n^2+4n+1}\right)\right)+\\ +\arctan\left(\frac{1}{2} \left(1+\sqrt{4 n^3 +4n^2+4n+1}\right)\right)$$ $\endgroup$
    – Raffaele
    Oct 29, 2017 at 20:16
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    $\begingroup$ Mathematica can find product,answer is enormus. $\endgroup$ Oct 29, 2017 at 20:18
  • $\begingroup$ @MariuszIwaniuk I am really interested in the answer ... ! $\endgroup$
    – Tolaso
    Oct 29, 2017 at 20:27

2 Answers 2

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We have $$S = \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^3 + n^2 + n + 1}\right) \\ = \Im\left(\ln\left(\prod_{n=1}^{\infty} \left(1 + \frac{i}{n^3+n^2+n+1}\right)\right)\right) \\ = \Im\left(\ln\left(\prod_{n=1}^{\infty} \frac{n^3+n^2+n+1+i}{n^3+n^2+n+1}\right)\right) \\ = \Im\left(\ln\left(\frac{\pi\operatorname{csch}(\pi)}{1+i}\prod_{k=1}^{3}\frac{1}{\Gamma(r_k)}\right)\right)$$ where $r_k$ is the $k$th root of $x^3-x^2+x-1-i$ (order doesn't matter since the product iterates over all of them). The last step comes from employing equation 19 on here (which comes directly from the Weierstrass factorization product formula for $\Gamma(x)$), and the fact that $\Gamma(i)\Gamma(-i) = \pi\operatorname{csch}(\pi)$.

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Something tells me this might be difficult to do by hand. Plugging in your product formula into Mathematica and then Simplifying, this is what comes out:

enter image description here

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    $\begingroup$ The arguments of $\Gamma$ look like roots of some cubic... $\endgroup$
    – Wojowu
    Oct 29, 2017 at 22:43
  • $\begingroup$ Shock !!! Where did that come from ? $\endgroup$
    – Tolaso
    Oct 29, 2017 at 23:08
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    $\begingroup$ Designating the arguments of $\Gamma$ as $a, b, c$ in order of appearance, $a+b+c = 4$, $ab + bc + ca = 6$, and $abc = 4 + i$. This gives $x^3 - 4x^2 + 6x - (4+i)$ as a minimal polynomial over $\mathbb{Z}[i]$ and $x^6 - 8x^5 + 28x^4 - 56x^4 + 68 x^2 - 48x + 17$ over $\mathbb{Z}$. WolframAlpha gives $a$ as one root of the cubic and decimals matching $b$ and $c$ for the others, so this is plausible. Of course, where the polynomials came from is anyone's guess... $\endgroup$ Oct 29, 2017 at 23:22
  • $\begingroup$ @ConnorHarris I'm guessing you meant $-56x^3$ there rather than $-56x^4$? :) $\endgroup$ Nov 26, 2017 at 1:03
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    $\begingroup$ @numbermaniac I did, yes. Good catch $\endgroup$ Nov 27, 2017 at 16:15

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