What does this infinite sum converge to?: $\sum_{n=1}^\infty \frac1{n^k} = \frac1{1^k} + \frac1{2^k} + \frac1{3^k} + ...$

Question

$$\sum_{n=1}^\infty \dfrac{1}{n^k} = \dfrac{1}{1^k} + \dfrac{1}{2^k} + \dfrac{1}{3^k} + \dfrac{1}{4^k} + \dfrac{1}{5^k} + ...$$

I've found that:

when $k=1$, it diverge to infinity

when $k=2$, it converge to $\dfrac{\pi^2}{6}$

when $k=4$, it converge to $\dfrac{\pi^4}{90}$

when $k=6$, it converge to $\dfrac{\pi^6}{945}$

when $k=8$, it converge to $\dfrac{\pi^8}{9450}$

when $k=10$, it converge to $\dfrac{\pi^{10}}{93555}$

when $k=12$, it converge to $\dfrac{691 \pi^{12}}{638512875}$

When $k$ tends to infinity, it seems to tend to $1$.

Though I didn't found anything for the cases $k=3$, $k=5$, $k=7$, $k=9$, $k=11$, etc...

I would like a formula that gives $\sum_{n=1}^\infty \dfrac{1}{n^k}$ according to $k$.

The formula needs to be as simple as possible.

Is the formula simpler when written in terms of $\pi$, $\eta = \dfrac{\pi}{2}$ or $\tau = 2 \pi$ ?

Answer 1

Perhaps you like the following formula for the Riemann zeta function. It is generalising the well-known formula for $\zeta(2n)$ in terms of Bernoulli numbers and powers of $\pi$ you have listed.

Answer 2

This is the Riemann Zeta function. There's simplified forms for even integers, but odd ones are much nastier.