Limit involving Zeta function

Question

Evaluate

$$\displaystyle \lim_{x \to 1} (x-1) \zeta (x)$$

I'm not much familiar with the properties of zeta function. An elementary solution is appreciated.

Thanks.

Answer 1

The detail of answer depends on how you define $\zeta(s)$, but if you are working with the definition

$$ \zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^s}, \quad s > 1 $$

then notice that the following inequality

$$ \frac{1}{s-1} = \int_{1}^{\infty} \frac{dx}{x^s} \leq \zeta(s) \leq 1 + \int_{1}^{\infty} \frac{dx}{x^s} = \frac{s}{s-1} $$

immediately yields

$$ \lim_{s \downarrow 1} (s-1)\zeta(s) = 1. $$

Answer 2

Let me expand on the hint given in this answer.

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The zeta function is the sum of the series $$ \zeta(s)=\sum_{n=1}^\infty\frac1{n^s} $$ The sum of the even terms is $$ 2^{-s}\zeta(s)=\sum_{n=1}^\infty\frac1{(2n)^s} $$ For any series, subtracting twice the sum of the even terms from the total sum gives the alternating sum: $$ \left(1-2^{1-s}\right)\zeta(s)=\sum_{n=1}^\infty(-1)^{n-1}\frac1{n^s} $$ Therefore, applying L'Hôpital, $$ \begin{align} \lim_{s\to1^+}(s-1)\zeta(s) &=\lim_{s\to1^+}\frac{s-1}{\left(1-2^{1-s}\right)}\lim_{s\to1^+}\sum_{n=1}^\infty (-1)^{n-1}\frac1{n^s}\\ &=\frac1{\log(2)}\sum_{n=1}^\infty(-1)^{n-1}\frac1n\\[6pt] &=1 \end{align} $$