Prove that this limit is the logarithmic derivative of the Riemann zeta function.

Question

Prove the following limit:

$$-\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}-\zeta (c)\right)$$

As a starting point I tried to enter this series expansion line in Mathematica:

Series[Zeta[s]*Zeta[c]/Zeta[s + c - 1] - Zeta[c], {c, 1, 0}]

which expanded at $c=1$ gives me:

$$-\frac{\zeta '(s)}{\zeta (s)}+O\left((c-1)^1\right)$$

But I don't know how to do series expansions.

Answer 1

Hint. assume $\zeta(s) \neq0$. One may recall that, as $c \to 1$, $$ (c-1)\zeta(c)=1+O(c-1) \tag1 $$ then, by the Taylor series expansion, as $c \to 1$, one has $$ \zeta(c+s-1)=\zeta(s)+\zeta'(s)(c-1)+O((c-1)^2) $$ giving $$ \begin{align} \frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}-\zeta (c)&=\left(\frac{\zeta (s)}{\zeta (c+s-1)}-1\right)\zeta (c) \\\\&=\left(\frac{\zeta (s)}{\zeta(s)+\zeta'(s)(c-1)+O((c-1)^2)}-1\right)\zeta (c) \\\\&=\left(\frac{1}{1+\frac{\zeta'(s)}{\zeta(s)}(c-1)+O((c-1)^2)}-1\right)\zeta (c) \\\\&=\left(-\frac{\zeta'(s)}{\zeta(s)}(c-1)+O((c-1)^2)\right)\zeta (c) \\\\&=-\frac{\zeta'(s)}{\zeta(s)}(c-1)\zeta (c)+O((c-1)^2\zeta (c)) \\\\& \to -\frac{\zeta'(s)}{\zeta(s)} \end{align} $$ as announced.