General P-series rule

Question

This is a p-series:

$$\sum_{n=1}^\infty \frac{1}{n^p}$$

There are 2 p-series (to my knowledge) that somehow reach pi:

$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ $$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$$

And I noticed that $\frac{\pi^3}{26}$ is really close to $\sum_{n=1}^\infty \frac{1}{n^3}$ and after a minute of digging on the internet I found this series which it's formula is $$ f(n)=(3^{n+1}+(-1)^{n+1})/2-2^{n+1}+1$$ And the series goes $0, 2, 6, 26, 90, 302, 966, 3026...$ and it seemed to line up with the p-series formula pretty well: $$\sum_{n=1}^\infty \frac{1}{n^2}=1.64453...$$ $$\frac{\pi^{2}}{f(2)}=1.64453...$$ $$$$ $$\sum_{n=1}^\infty \frac{1}{n^3}=1.20205...$$ $$\frac{\pi^{3}}{f(3)}=1.19254...$$ $$$$ $$\sum_{n=1}^\infty \frac{1}{n^4}=1.08232...$$ $$\frac{\pi^{4}}{f(4)}=1.08232...$$ $$$$ $$\sum_{n=1}^\infty \frac{1}{n^5}=1.03692...$$ $$\frac{\pi^{5}}{f(5)}=1.01331...$$ $$$$ $$\sum_{n=1}^\infty \frac{1}{n^6}=1.01734...$$ $$\frac{\pi^{6}}{f(6)}=0.99522...$$

But sadly that certain function only lines up perfectly with $p=2$ and $p=4$ but I was wondering if someone here can modify the function $f(n)$ so: $$\frac{\pi^{x}}{f(x)} = \sum_{n=1}^\infty \frac{1}{n^x}$$

Edit: $f(x)$ starting from 2 should look somthing like: $$6, 25.794, 90, 295.122, 945, 2995.285, 9450...$$

Helpful link: https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function — Oleg567, Jul 12 '18 at 15:53
This answer might be useful, at least in the cases of the even exponents. — robjohn, Jul 12 '18 at 15:55
See also https://math.stackexchange.com/questions/248413/zeta-function-values-in-terms-of-bernoulli-numbers. — Batominovski, Jul 12 '18 at 15:59
Seens like every $2k$-serie has the form $\frac{\pi^{2k}}{n}$. I actually don't know how to express this $n$ in function of $k$. For curiosity, $\sum\frac{1}{n^{6}}=\frac{\pi^{6}}{945}, \sum\frac{1}{n^{8}}\frac{\pi^{8}}{9450}$ and $\sum\frac{1}{n^{10}}\frac{\pi^{10}}{93555}$ — Mateus Rocha, Jul 12 '18 at 16:15
@MateusRocha Yes after trying some numbers I think every even number is whole and every odd number is irrational... I'll try some more numbers and write the series that $f(x)$ needs to look like — Amit TPB, Jul 12 '18 at 16:18
Actually that make sense. Take a look in this equation: $$\zeta(2n)=(-1)^{n+1}\frac{B_{2n}(2n)^{2n}}{2(2n)!}. $$ Here $\zeta(s)=\sum\frac{1}{n^{s}}$ and $B_{n}$ is the n-Bernoulli number — Mateus Rocha, Jul 12 '18 at 16:29
@MateusRocha: shouldn't there be a $\pi^{2n}$ in that formula? — robjohn, Jul 12 '18 at 16:30
You are right. The correct is $$\zeta(2n)=(-1)^{n+1}\frac{B_{2n}(2\pi)^{2n}}{2(2n)!} $$ — Mateus Rocha, Jul 12 '18 at 17:02

score 3 · Answer 1 · answered Jul 12 '18 at 17:05

What you're calling a p-series is what mathematicians call the Riemann zeta function. You will find that for any even power, there is a power of $\pi$ in the result:

$\zeta(2n) = (-1)^{n+1}\frac{B_{2n}(2 \pi)^{2n}}{2(2n)!})$

where $B_k$ denotes the $k$th Bernoulli number.

There's much more at https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function

For the odd positive integers, much less is known. It is known that $\zeta(3)$ is irrational, that at least one of $\zeta(5),\ \zeta(7),\ \zeta(9),\ \zeta(11)$ is irrational, and that there are infinitely many irrational values for $\zeta(2n+1)$. But no simple closed-form solution of the type you imagine is known.

score 0 · Answer 2 · answered Jul 12 '18 at 17:41

What does happen to be a rational multiple of $\pi^n$ is $$ \sum_{k=0}^\infty\frac{(-1)^{kn}}{(2k+1)^n}=\left\{\begin{array}{}\beta(n)&\text{if $n$ is odd}\\\zeta(n)\left(1-2^{-n}\right)&\text{if $n$ is even}\end{array}\right. $$ A recursion for the Dirichlet beta function, $\beta(n)$, is given for odd $n$ in this answer.

A recursion for the Riemann zeta function, $\zeta(n)$, is given for even $n$ in this answer.

General P-series rule

2 Answers2