Alternating series of 'zeta' functions for real integer parameters equals 1/2

Question

I would like to know how to show that:

$$ \sum_{s=2}^{\infty}(-1)^s\zeta(s)=\dfrac{1}{2} $$

where $$ \zeta(s)=\sum_{n=1}^{\infty}\dfrac{1}{n^s} \\s,n \in N $$

Kind regards

Answer 1

While technically the sum does not converge, you can get some sort of regularization. Start by using an integral representation of the zeta function:

$$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x -1}dx$$

Your sum can then be re-written as

\begin{align} \sum_{s=2}^{\infty}(-1)^s\zeta(s)&= \sum_{s=2}^{\infty} (-1)^s \frac{1}{(s-1)!}\int_0^\infty \frac{x^{s-1}}{e^x -1}\,dx \\ &= \int_0^\infty \frac{1}{e^x-1} \sum_{s=1}^{\infty} \frac{(-1)^{s+1}}{s!} x^s \,dx \\ &=-\int_0^\infty \frac{1}{e^x-1} \sum_{s=1}^{\infty} \frac{(-x)^s}{s!}\,dx \\ &= -\int_0^\infty \frac{e^{-x}-1}{e^x-1}\,dx \\ &= \int_0^\infty \frac{1-e^{-x}}{e^x-1}\,dx \\ &=\int_0^\infty \frac{dx}{e^x-1}-\int_0^\infty \frac{e^{-x}}{e^x-1}\,dx \\ &=\int_0^\infty \frac{e^{-x}}{1-e^{-x}}\,dx-\int_0^\infty \frac{e^{-2x}}{1-e^{-x}}\,dx \\ &= \int_0^\infty \sum_{n=0}^\infty e^{-(n+1)x}\,dx - \int_0^\infty \sum_{n=0}^\infty e^{-(n+2)x}\,dx \\ &=\sum_{n=0}^\infty \frac{1}{n+1}-\frac{1}{n+2} \\ &=1 \end{align} Where the last sum can be evaluated, for example, by telescoping. Of course, these equals signs should be regarded in a weak sense, as the sum doesn't formally converge, but alternates between $1.5$ and $0.5$.