# Thread: A limit on Riemann zeta function

1. ## A limit on Riemann zeta function

Hi! I just found a limit (maybe someone has found it before) experimentally that
$
\displaystyle \lim_{s\to 1} \left (1-x^{1-s}\right )\zeta(s)=\ln x.
$

But I can't prove it. Can anyone show a proof to me? Thanks!

2. I think I have it. Since we are talking about the complex plane we better have a definition of the zeta function in a neighborhood around s = 1. So from mathworld, a series representation is:

$\displaystyle \zeta (s) = \frac{1}{1-2^{1-s}}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$

So now the above becomes:

$\displaystyle (1-x^{1-s})\zeta (s) = \frac{1-x^{1-s}}{1-2^{1-s}}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$

From wikipedia we have that

$\displaystyle \log(z) = \sum_{n=1}^\infty -\frac{(1-z)^n}{n}$ and for z = 2 we have

$\displaystyle \log(2) = \sum_{n=1}^\infty -\frac{(-1)^{n}}{n} = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}$

Hence now we have

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$\displaystyle = \underset{s \to 1}{\lim} \left(\frac{1-x^{1-s}}{1-2^{1-s}}\right) \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s} = \ln(2) \;\; \underset{s \to 1}{\lim} \;\frac{1-x^{1-s}}{1-2^{1-s}}$

I factored out the sum from the limit since at its limiting value it is just the logarithm of 2 so it does not affect the limit. By L'Hopital's rule the limit is:

$\displaystyle \underset{s \to 1}{\lim} \;\frac{1-x^{1-s}}{1-2^{1-s}} = \underset{s \to 1}{\lim} \;\frac{-x^{1-s}\ln(x)}{-2^{1-s}\ln(2)} = \frac{\ln(x)}{\ln(2)}$

Multiplying things out you get the desired expression.

3. $\displaystyle{ \lim_{s \to 1 } ( 1 - x^{1-s} ) \zeta(s) }$

$= \displaystyle{ \lim_{s \to 1 } \frac{1 - x^{1-s}}{s-1} [(s-1) \zeta(s) ] }$

$= \displaystyle{ \lim_{s \to 1 } \frac{1 - x^{1-s}}{s-1} \cdot \lim_{s \to 1 } (s-1)\zeta(s)$

$= \ln(x) \cdot \displaystyle{ \lim_{s \to 1 } (s-1)\zeta(s)}$

Consider the area under the curve $f(x) = \frac{1}{x^s} ~,~ s>1$ from $x = 1$ to $x= n+1$ and the rectangles with dimensions $1 \times f(1) ~,~ 1\times f(2) ~,~ ...~,~ 1\times f(n)$ .

We have this inequality :

$\int_1^n f(x) ~dx + 1 > f(1) + f(2) + f(3) + ... + f(n) > \int_1^{n+1} f(x) ~dx$ as $f(x)$ is decreasing for $x > 0$ .

Therefore , $s - L(n) > (s-1) H_n > 1 - L(n+1) ~~,~~ L(n) = \frac{1}{n^{s-1} }$

When $n$ tends to infinity , we have

$s > (s-1) \zeta(s) > 1 ~~,~~ s>1$ ( $\displaystyle{\lim_{n\to\infty} L(n) = 0 } )$

$\displaystyle{ \lim_{s \to 1+ } s > \lim_{s \to 1^+ } (s-1)\zeta(s) > \lim_{s \to 1^+ } 1 }$

But $\displaystyle{ \lim_{s \to 1^+ } s = 1 }$ which forces $\displaystyle{ \lim_{s \to 1^+ } (s-1)\zeta(s) = 1 }$ . Therefore , $\displaystyle{ \lim_{s \to 1 } ( 1 - x^{1-s} ) \zeta(s) } = \ln(x)$

4. I like your proof. Here is a shorter version of the proof of the limit. Consider the Laurent series of zeta around s =1 :

$\displaystyle \zeta(s)=\frac{1}{s-1}+\sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n \; (s-1)^n$.

$\displaystyle (s-1)\zeta(s)= 1+\sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n \; (s-1)^{n+1}$.

Taking the limit as s goes to 1 is now a piece of cake. Although if you didn't have the Laurent series, it'd be more difficult to show.

However, if we are aware of the fact that s-1 is a simple pole, we will immediately get a Laurent series like the above, valid is some neighborhood around s=1 where we don't really care about the coefficients. Then we have the same situation as above.

EDIT: I just realized that interchanging the limit and the sum will require justification. That is we need the series to be uniformly convergent.