TITLE: Why doesn't this change of variables work?
QUESTION [1 upvotes]: I would like to calculate the infinite sum of odd terms of $\zeta(2)$ as follows. Setting $2k-1 = \ell$ yields $k = (\ell+1)/2$ and so
$$\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}} = \sum_{\frac{l+1}{2}=1}^{\infty}\frac{1}{\ell^{2}}  = \sum_{\ell = 1}^{\infty} \frac{1}{\ell^{2}} = \frac{\pi^{2}}{6}.$$
This is obviously nonsense if you instead try splitting up $\zeta(2)$ into its even and odd terms.
Experience shifting indices without problem seems to suggest that it's the division by 2 that is causing the issue, but its not obvious to me what is wrong. Is there an abuse of notation when using the $=$ symbol in the infinite sum that is making this operation invalid or is there something else going on?

REPLY [3 votes]: $$\sum_{\frac{\ell+1}{2}=1}^{\infty}\frac{1}{\ell^{2}}  = \sum_{\ell = 1}^{\infty} \frac{1}{\ell^{2}}$$
is the problem. The left is perfectly valid if you solve the equations
$$\frac{\ell+1}{2}=1\qquad\frac{\ell+1}{2}=2\qquad\frac{\ell+1}{2}=3\qquad\ldots$$for $\ell$. But you substituted $\ell$ in the place of $\frac{\ell+1}{2}$ on the right which gives the equations
$$\ell=1\qquad\ell=2\qquad\ell=3\qquad\ldots$$
Though the first equation may give the correct value for $\ell$, the remaining ones do not. Thus you get the wrong answer because
$$\ell \neq \frac{\ell+1}{2}$$
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