Derivation of $\lim_{s\to1}\zeta(s)-\log\prod_{n=1}^\infty(1+n^{-s})=\gamma$

Question

I want to prove that$$\lim_{s\to1}\zeta(s)-\log\prod_{n=1}^\infty(1+n^{-s})=\gamma$$

I know that for $|s|>0$, $\log(1+n^{-s})=\sum_{k=1}^\infty (-1)^{k+1}\frac{n^{-sk}}{k}$

So $$\log\prod_{n=1}^\infty(1+n^{-s})=\sum_{n=1}^\infty\sum_{k=1}^\infty(-1)^{k+1}\frac{n^{-sk}}{k}$$

If I switch the sums: $$\log\prod_{n=1}^\infty(1+n^{-s})=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}\sum_{n=1}^\infty n^{-sk}=\sum_{k=1}^\infty\frac{(-1)^{k+1}\zeta(ks)}{k}$$

The rest follows from $$\gamma=\sum_{k=2}^\infty\frac{(-1)^k\zeta(k)}{k}$$

My question is if there are any Theorems or any other proof for the justification of switching the sums?

Answer 1

It is easy to justify the interchange of the sums. Use the following: $\sum n^{-k/2} \leq \int _1 ^{\infty} x^{-k/2} dx =\frac 1 {k/2 -1}$ for $k>3$ and $\sum \frac 1 {k(k/2-1)} < \infty$. Hence the double sum is absolutely convergent which allows us to interchange the order of summation.