the definition of Riemann zeta function

Question

I've read that the Riemann zeta function for $0<s<1$ is defined :

$$\lim_{x\rightarrow\infty} \left(\sum_{n \leq x}\frac{1}{n^s}- \frac{x^{1-s}}{1-s}\right)$$

I don't know how to prove that this limit exists.

help me please.

thanks a lot.

What have you done so far? And are you considering the complex plane or just the real numbers? — graydad, Oct 16 '14 at 20:05
Riemann-Zeta function is not defined on the real numbers for $0<s \leq 1$. You need $s>1$ — graydad, Oct 16 '14 at 20:07
see the book "introduction to analytic number theory" by Apostol page 55.it is there. — user115608, Oct 16 '14 at 20:08
The proof is in Apostol on the following pages. Essentially, you use the Euler Maclaurin formula (which is given earlier in the same chapter) to turn the sum into an integral, then you bound the pieces. — J. David Taylor, Oct 16 '14 at 20:22
@graydad: The (real) Euler zeta function is defined by just the basic series on the reals greater than $1$. The Riemann zeta function, defined by analytic continuation of the Euler zeta function into the complex plane with the point $1$ removed, is well defined on the real interval $(0,,1)$. — John Bentin, Oct 16 '14 at 21:25
@Jonathan Taylor i,ve seen Apostol.there's no proof.help please. — user115608, Oct 17 '14 at 02:51

score 1 · Answer 1 · answered Oct 17 '14 at 05:59

Pg 55 of Apostol, theorem 3.2 (b) (halfway down page 56 he gives the proof) gives

$\sum_{n\leq x}\frac{1}{n^s}=\frac{x^{1-s}}{1-s} +\zeta(s) +O(x^{-s})$

if $1\neq s>0$ Subtract the first term of the right hand side and take $x$ to infinity.

(and the proof does consist of Euler Summation)

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