# Is this the correct closed form for a series similar to $\zeta(2)$?

I hope this question is well received. I don't have a computer that can calculate very many terms for the infinite series: $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(2n-1)^{2}},$$ but is it going to equal to this closed form: $\log(2.5)$?

The sum is equal to Catalan's constant, which is close but not equal to $\log(5/2).$ The difference between the two is $-0.000325138$
• @user124808 At least it's not known to -- the sum you've given is the usual definition of Catalan's constant. But it is an open problem even to determine whether $G$ is a rational number. – David Zhang May 11 '18 at 20:47
• @GeraldEdgar, since this series is the $L$-function of the nontrivial character mod $4$, which is an odd character (the trivial character, which is even, is related to the zeta-function), the value should be thought of as like odd zeta-values rather than even zeta-values, so we shouldn't expect the series to have any simple relation to powers of $\pi$. For comparison, $L(1,\chi_4) = \pi/4$ and $L(3,\chi_4) = \pi^3/32$ are rational multiples of suitable powers of $\pi$. – KConrad May 11 '18 at 23:55