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It's easy to prove that$$\zeta(1)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...$$ diverges, and $$\zeta(2)=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...$$ converges to $\frac{\pi^2}{6}$.

Intuiting the result, this means that there exists a real $x$, $1<x<2$, such that it is the minimum $x$ satisfying $\zeta(x)=L$ (i.e. $\zeta(x)$ converges to a limit). What is the value of $x$?

Similarly, is it known which $x$ (again $1<x<2$) is the minimum of $$\sum_{\text{p prime}}\frac{1}{p^x}=L'$$ and must $L'=L$?

Meow
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    I'll take exception on the "easy to prove" $\zeta(2)$... – vonbrand Apr 18 '13 at 16:12
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    @vonbrand ...as long as Euler's metaphorical shoulders exist. – Meow Apr 18 '13 at 16:13
  • @vonbrand : I think $\zeta(2)$ can be computed easily using the Fourier series of a function with a jump dicontinuity, but I don't remember exactly what that function is. – Stefan Smith Apr 18 '13 at 23:07
  • @StefanSmith, I've seen my fill of (wonderful, absolutely non-obvious) proofs, thank you... [got me intrigued, anyway... OK, will formalize as a question]. – vonbrand Apr 18 '13 at 23:18

1 Answers1

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There is no minimum $x$. $$f(x) = 1 + \dfrac1{2^x} + \dfrac1{3^x} + \cdots + \dfrac1{n^x} + \cdots$$ converges for $x > 1$ and diverges for $x \leq 1$. To put it the other way around, the maximum value of $x$ for which $f(x)$ diverges is $x=1$.

EDIT

First note that we have $\dfrac1{k^x}$ is a decreasing function of $x$. Hence, for $x \leq 1$ we have $$\sum_{k=1}^n \dfrac1{k^x} \geq \sum_{k=1}^n \dfrac1k$$ Hence, $\displaystyle \sum_{k=1}^{\infty} \dfrac1{k^x}$ diverges for $x \leq 1$.

To prove convergence for $x > 1$, see the following question:

Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p > 1$

EDIT

The same is true for the primes as well, i.e., $$\sum_{p \in \text{ primes}} \dfrac1{p^x}$$ diverges for $x \leq 1$ and converges for $x>1$.

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