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J
Thanks a lot. This is a derivation that I can follow. For physics purposes it is of course much easier to look up the zeta function and...
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J
The particular derivation I gave requires knowledge of complex analysis and is quite lengthy. I therefore thought that it illustrated...
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J
Extracting relevant bits from other thread: We can write
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \sum_{n=1}^\infty...
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J
In another thread I provide a derivation of the general closed formula for ##\zeta(2k)## in terms of Bernoulli numbers:
\begin{align*}...
Like.
Thanks a lot. This is a derivation that I can follow. For physics purposes it is of course much easier to look up the zeta function and...