<s>[INST] Calculating squared reciprocals of arithmetic series Let $n>0$ be an integer. Is it possible to calculate the value of the sum
$$1+\frac1{(1+n)^2}+\frac1{(1+2n)^2}+\ldots$$?
 [/INST] One can prove Marko Riedel's claim using a simple Taylor series argument. We have
$$
\sum_{k=1}^\infty \frac{1}{(kn+1)^2} = \sum_{k=1}^\infty \frac{1}{(kn)^2} \left(\frac{1}{(1+(kn)^{-1})^2}\right),
$$
as well as the Taylor series
$$
\frac{1}{(1+x)^2} = 1 - 2x + 3x^2 - 4x^3 + \cdots,
$$
absolutely convergent for $|x| < 1$. The right hand side of the penultimate display is
$$
\sum_{k=1}^\infty \frac{1}{(kn)^2} \left(1 - \frac{2}{kn} + \frac{3}{(kn)^2} - \frac{4}{(kn)^3} + \cdots \right) = \sum_{k=1}^\infty \sum_{q=2}^\infty \frac{(q-1)(-1)^q}{(kn)^q}.
$$
Since the Taylor series is absolutely convergent, we may interchange the order of summation. The sum over $k$ gives $\zeta(q)$, which yields Riedel's claim.
In the special case, $n = 2$, we can exploit a bit of symmetry to obtain
$$
\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{4n^2} = \frac{3}{4}\zeta(2) = \frac{\pi^2}{8}.
$$
</s>