I write $\zeta(s)$ for the Riemann zeta function. I denote Riemann's xi-function by $\xi(s).$ Here $s$ is a a whole number greater than or equal to zero.
Problem: Show that $$ \pi=\log\left(-\sum\limits_{s=0}^{\infty}(1-s)\frac{\zeta(s)}{\xi(s)}\right). $$
My Attempt At A Solution: This is just an exercise in substitutions. Following Ahlfors Chapter 5.4.3 he derives Riemann's xi(s)-function, denoted $\xi(s)$ as: $$ \xi(s)=(1-s)\left[\frac{(s/2)\Gamma(s/2)}{\pi^{s/2}}\right]\zeta(s) $$ On the other hand I have the following equation for the volume of the unit-ball in $s-$dimensional Euclidean space $$ \frac{\pi^{s/2}}{(s/2)\Gamma(s/2)}. $$ The bracketed term of Riemann's xi function is simply the inverse of the volume of the $s-$ball, which for short handedness I will denote by $\nu(s).$ After substitution I write $$ \xi(s)=(1-s)\frac{\zeta(s)}{\nu(s)}. $$ Subsequently, $$ \frac{\xi(s)}{(1-s)\zeta(s)}=\frac{1}{\nu(s)}. $$ Of course at $s=1$, we have a pole at $\zeta(s).$ However in this case $1-s=0$ and so we have a cancellation.
Now, whenever $s\equiv 0\pmod2,$ $ \sum_{s=0}^{\infty}\nu(s)=e^{\pi}. $ And so, whenever $s$ is an even number: $$ e^{\pi}=-\sum\limits_{s=0}^{\infty}(1-s)\frac{\zeta(s)}{\xi(s)} $$ Note the content of the summation is negative and the negative of a negative is positive and so we can take our logarithm, $$ \pi=\log\left(-\sum\limits_{s=0}^{\infty}(1-s)\frac{\zeta(s)}{\xi(s)}\right). $$ I checked my solution using Wolfram Alpha: check.