Calculating squared reciprocals of arithmetic series Let $n>0$ be an integer. Is it possible to calculate the value of the sum
$$1+\frac1{(1+n)^2}+\frac1{(1+2n)^2}+\ldots$$?
 A: What follows is not a closed form but a different representation. The series part of the sum is
$$S(x) = \sum_{k\ge 1} \frac{1}{(1+kx)^2}$$
evaluated at $x=n.$
The sum term is harmonic and may be evaluated by inverting its Mellin transform.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = 1, \quad \mu_k = k \quad \text{and} \quad
g(x) = \frac{1}{(1+x)^2}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty \frac{1}{(1+x)^2} x^{s-1} dx
= \left[- \frac{1}{1+x} x^{s-1} \right]_0^\infty
+ (s-1) \int_0^\infty \frac{1}{1+x} x^{s-2} dx.$$
The bracketed term vanishes for $1<\Re(s)<2.$ 
Now the Mellin transform $h^*(s)$ of $h(x) = \frac{1}{1+x}$ is easily seen to obey
(contour integration with a keyhole contour, slot on the positive real axis)
$$h^*(s) (1-e^{2\pi i (s-1)}) = 2\pi i\times
\mathrm{Res}\left(\frac{1}{1+x} x^{s-1} ; x=-1 \right)$$
which gives 
$$h^*(s) = \frac{2\pi i}{1-e^{2\pi i s}} e^{i\pi (s-1)}
= - \frac{2\pi i}{1-e^{2\pi i s}} e^{i\pi s}
= - \pi \frac{2i}{e^{-i\pi s}-e^{i\pi s}}
= \frac{\pi}{\sin(\pi s)}.$$
Hence the Mellin transform $g^*(s)$ of $g(x)$ has the form
$$g^*(s) = (s-1) \frac{\pi}{\sin(\pi (s-1))}
= (1-s) \frac{\pi}{\sin(\pi s)}.$$
The harmonic sum identity then implies that the Mellin transform $Q(s)$ of $S(x)$ is given by
$$(1-s) \frac{\pi}{\sin(\pi s)} \zeta(s)
\quad\text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \zeta(s).$$
The Mellin inversion integral for this case is
$$\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds.$$
Now inverting the transform in the right half plane to obtain an expansion about infinity, we have that from the poles at $q\ge 2$ the sum of their residues is
$$S(x) = -\sum_{q\ge 2} (1-q) (-1)^q \zeta(q) / x^q.$$
This produces for the original sum the representation
$$1+ \sum_{q\ge 2} (q-1) (-1)^q\zeta(q) \frac{1}{n^q},$$
convergent for $n>1.$
A: For $n > 0$ (whether integer or not is irrelevant)
$$\sum_{j=0}^\infty \dfrac{1}{(1+jn)^2} = \dfrac{1}{n^2} \Psi'(1/n)$$
where $\Psi$ is the Digamma function.  Known values include
$$ \eqalign{n = 1, & sum = \dfrac{\pi^2}{6}\cr
            n = 2, & sum = \dfrac{\pi^2}{8}\cr
            n = 4, & sum = \dfrac{\pi^2}{16} + \dfrac{{\rm Catalan}}{2}\cr}$$
A: What about
$$\sum_{n=0}^\infty\sum_{j=0}^\infty \dfrac{1}{(1+jn)^2} = \sum_{n=0}^\infty \dfrac{1}{n^2} \Psi'(1/n)?$$
A: 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
$$
\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{1}{4n^2} = \frac{3}{4}\zeta(2) = \frac{\pi^2}{8}.
$$
