I have a lot of sum questions right now ... could someone give me the convergence of, and/or formula for, $\sum_{n=2}^{\infty} \frac{1}{n^k}$ when $k$ is a fixed integer greater than or equal to 2? Thanks!!
P.S. If there's a good way to google or look up answers to these kinds of simple questions ... I'd love to know it...
Edit: Can I solve it by integrating $\frac{1}{x^k}$ ? I can show it converges, but to find the formula? Is my question just the Riemann Zeta function?
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Thanks guys! This got me the following result:
$\sum_{p} \frac{1}{p} \log \frac{p}{p-1} ~ ~ \leq ~ \zeta(2)$
summing over all primes $p$. (And RHS is Riemann zeta function.)
First, sum over all integers $p$ instead of primes. Then transform the log into $\sum_{m=1}^{\infty} \frac{1}{m p^{m}}$ (reference: wikipedia. I know.). Now we have (with rearranging):
$\leq ~ \sum_{m=1}^{\infty} \frac{1}{m} \sum_{p=2}^{\infty} \frac{1}{p^{m+1}}$
By the result of this question (Arturo's answer), this inner sum, which is $\zeta(m+1)$ is at most $\frac{1}{m+1-1} = \frac{1}{m}$. So we have
$\leq ~ ~ \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{m} = \zeta(2)$
I think this is a very pretty little proof. Thanks again, hope you math people enjoyed reading this....