4
$\begingroup$

Can you provide a proof for at least one of the claims given below?

It is known that $\pi=\displaystyle\sum_{n=1}^{\infty}\frac{3^n-1}{4^n} \cdot \zeta(n+1)$ where $\zeta$ denotes Riemann zeta function (Vardi 1991) . Similarly we can formulate the following claims:

Claim 1. $$\frac{\pi}{\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{2^n-1}{3^n} \cdot \zeta(n+1)$$

The SageMath cell that demonstrates this claim can be found here.

Claim 2. $$\sqrt{3} \pi=\displaystyle\sum_{n=1}^{\infty}\frac{5^n-1}{6^n} \cdot \zeta(n+1)$$

The SageMath cell that demonstrates this claim can be found here.

Claim 3. $$(\sqrt{2}+1) \pi=\displaystyle\sum_{n=1}^{\infty}\frac{7^n-1}{8^n} \cdot \zeta(n+1)$$

The SageMath cell that demonstrates this claim can be found here.

Claim 4. $$(\sqrt{3}+2) \pi=\displaystyle\sum_{n=1}^{\infty}\frac{11^n-1}{12^n} \cdot \zeta(n+1)$$

The SageMath cell that demonstrates this claim can be found here.

$\endgroup$
4
  • 1
    $\begingroup$ I might be wrong... these can be obtained from manipulating the parametrized sums from the same wikipedia article en.wikipedia.org/wiki/Riemann_zeta_function#Infinite_series . One takes certain linear combinations of sums to obtain your series, while on the RHS the polygamma functions will simplify due mirror symmetry. This is not the right forum for this question. $\endgroup$
    – Nemo
    Commented Aug 11, 2021 at 5:49
  • $\begingroup$ Even simpler is to use this generating function en.wikipedia.org/wiki/Polygamma_function#Taylor_series $\endgroup$
    – Nemo
    Commented Aug 11, 2021 at 10:19
  • 4
    $\begingroup$ Nemo, I would love to see some details (it does not seem obvious or simple to me)! $\endgroup$ Commented Aug 11, 2021 at 13:29
  • 1
    $\begingroup$ It is interesting that $1+\sqrt{2}$ is a unit (fundamental?) in the real quadratic field with discriminant 8, $2+\sqrt{3}$ is a unit (fundamental?) in the real quadratic field of discriminant 12. $\endgroup$
    – Stopple
    Commented Aug 11, 2021 at 15:30

0

You must log in to answer this question.