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Let $a > 0$ et $f$ defined by $$ f(a) = \sum_{k = 1}^{+ \infty} \frac{1}{1 + (ak)^2} $$ We have

$$ \exists ! a > 0, \sum_{k = 1}^{+ \infty} \frac{1}{1 + (ak)^2} = \frac{1}{2} $$

I want to find $a$. So, I tried several thing :

1 - We know $\frac{1}{1 + x^2} = \arctan'(x)$, so

$$ \frac{1}{2} = f(a) = \sum_{k = 1}^{+ \infty} \arctan'(ak) = \sum_{k = 1}^{+ \infty} \frac{1}{\tan'(\arctan(ak))} $$

because $\arctan(y)' = \frac{1}{\tan'( \arctan(y) )}$. Now $\tan'(x) = \frac{1}{\cos^2(x)}$ so we have to find $a$ s.t

$$ \frac{1}{2} = \sum_{k = 1}^{+ \infty} \cos^2(\arctan(ak)) $$

But I don't how to deal with this sum.

2 - With a computer, I found that $a$ is close to $1.644$ and I conjecture that $a$ could be equal to a number like $ \frac{\pi^2}{6} $. But

$$ f(\frac{\pi^2}{6} ) \approx 0.4977 $$

I guess that $a$ may be a limit of the zeta function. So, I have to prove the following implication

$$ \sum_{k = 1}^{+ \infty} \frac{1}{t^2} = a \Rightarrow \sum_{k = 1}^{+ \infty} \frac{1}{1 + (ak)^2} = \frac{1}{2} $$

But I am not sure if it works.

3 - I guess that we may found $a$ like we prove $ \sum \frac{1}{n^2} = \frac{\pi^2}{6} $. But I am not sure if it's work.

Now I am in lack of idea.. so if anyone can give me a hint, I appreciate your help.

Tohiea
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  • Are you sure of the $1.644$? I see the answer as being just slightly less than $1.64$ – lulu Nov 21 '20 at 23:06
  • @lulu I am not sure... When I compute with $a = 1.644$ and $N = 1000000$ term I found $$ \sum_{k = 1}^N \frac{1}{1 + (ak)^2} = 0.4982314373690213 \approx 0.5 $$ – Tohiea Nov 21 '20 at 23:09
  • Yeah..on working with greater precision, I agree that the answer is a hair greater than $1.64$. With $a=1.6405$ and $N=10^8$ I get $0.50000582$ – lulu Nov 21 '20 at 23:30
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    A tiny bit larger than $1.64$ sounds correct. The analytical solution would be $a = \frac{\pi}{y}$ where $y\coth(y) = 2$ which gives $\approx 1.64051$. See e.g. https://math.stackexchange.com/questions/991066/method-of-proof-of-sum-limits-n-1-infty-tfrac-coth-n-pin7-tfrac19/991110#991110 for how to evaluate the sum – Winther Nov 21 '20 at 23:31
  • @lulu Yes, I agree – Tohiea Nov 21 '20 at 23:46
  • @Winther Yes thank you, it is exactly what I need to pursue. – Tohiea Nov 21 '20 at 23:48
  • I updated my answer – Claude Leibovici Nov 23 '20 at 03:52

1 Answers1

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$$f(a) = \sum_{k = 1}^{+ \infty} \frac{1}{1 + (ak)^2}=\frac{\pi \coth \left(\frac{\pi }{a}\right)}{2 a}-\frac{1}{2}$$ Let $x=\frac \pi a$ and the problem is to find the zero of function $$g(x)=x \coth(x)-2$$

Using Taylor series $$x \coth(x)=1+\frac{x^2}{3}-\frac{x^4}{45}+\frac{2 x^6}{945}-\frac{x^8}{4725}+\frac{2 x^{10}}{93555}+O\left(x^{12}\right)$$ and series reversion gives $$x=t+\frac{t^3}{30}+\frac{t^5}{1400}-\frac{t^7}{42000}-\frac{163 t^9}{77616000}+O\left(t^{11}\right)$$ with $t=\sqrt{3(g(x)+1)}$.

Using $g(x)=0$ as desired leads to $$x=\frac{9534829 }{8624000}\sqrt{3} \implies a=\frac{8624000 \pi }{9534829 \sqrt{3}}\approx 1.640533$$ while the solution given by Newton method is $1.640511$.

Edit

To show the impact of the level of expansion of the result, consider the expansion $$x \coth(x)=\sum_{n=0}^{2p}\frac{2^{2 n} B_{2 n}}{(2 n)!}x^{2p}+O(x^{2p+2})$$ and use series reversion to obtain $$\left( \begin{array}{cc} n & a_{(n)} \\ 1 & 1.813799364 \\ 2 & 1.648908513 \\ 3 & 1.639328024 \\ 4 & 1.640281061 \\ 5 & 1.640533429 \\ 6 & 1.640521573 \\ 7 & 1.640511406 \\ 8 & 1.640511008 \\ 9 & 1.640511422 \\ 10 & 1.640511481 \\ 11 & 1.640511466 \\ 12 & 1.640511462 \end{array} \right)$$

Claude Leibovici
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  • Thank you for you help. Some question, we get $x = \frac{ 9534829 }{ 8624000 } \sqrt{3} $ because $$ g(x) = 0 \Rightarrow t = \sqrt{3} $$ Then, with this value, we evaluate in $x = t + \frac{t^3}{30} + \cdots = \sqrt{3} + \frac{\sqrt{3}^3}{30} + \cdots $, it is correct ? – Tohiea Nov 22 '20 at 11:19
  • Also, you write $x \coth(x)$ as a Taylor series at order 12, the method will always work if we write $x \coth(x)$ at order 7? thank you. – Tohiea Nov 22 '20 at 11:40
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    @Tohiea. Correct for both questions. More terms you use and better accuracy. I shall try tomorrow to add numbers. – Claude Leibovici Nov 22 '20 at 16:05