Finding the value of $\sum^{\infty}_{n=3,5,7,9....}\frac{2n^2 \exp(-\pi n/2)}{\exp(\pi n)+1}$ I want to evalutate:
$\displaystyle \tag*{} \sum \limits ^{\infty}_{n=3,5,7,9....}\dfrac{2n^2 \exp \left(-\pi n/2\right)}{\exp(\pi n)+1}$
This question is inspired from my previous question which again was inspired from quora, while the sum with $n$ in numerator is difficult to find closed form, however $n^2$ is easy to find. Any help would be appreciated.

Let us write $q=e^{-\pi/2}$ and the sum in question becomes $$F(q)=2\sum_{n\text{ odd}, n>1}\frac{n^2q^{3n}}{1+q^{2n}}\tag{1}$$ which can be further expressed as $$F(q) =2\sum_{n\text{ odd}, n>1}n^2q^n-2\sum_{n\text{ odd}, n>1}\frac{n^2q^n}{1+q^{2n}}\tag{2}$$ The first sum is easy to handle and we can just note that $$\sum n^2q^n=\left(q\frac{d}{dq}\right)^2\frac{1}{1-q}=\frac{q+q^2}{(1-q)^3}=a(q) \text{ (say)} $$ and hence $$A=2\sum_{n\text { odd}, n>1}n^2q^n=2(a(q)-4a(q^2)-q)=2q^3\cdot\frac{9-2q^2+q^4}{(1-q^2)^3}\tag{3}$$ Let $$g(q) =\sum_{n=1}^{\infty}\frac{n^2q^n}{1+q^{2n}}$$ then the second sum, say $B$, in $(2)$ can be expressed as $$B=2g(q)-8g(q^2)-\frac{2q}{1+q^2}$$ To evaluate $g(q) $ in closed form we need a bit of elliptic function theory. The function $\operatorname {dn} (u, k) $ has the Taylor series expansion $$\operatorname {dn} (u, k) =1-k^2\frac{u^2}{2!}+k^2(4+k^2)\frac{u^4}{4!}+\dots$$ and it also has a Fourier series $$\operatorname {dn} (u, k) =\frac{\pi} {2K}+\frac{2\pi}{K}\sum_{n=1}^{\infty} \frac {q^n} {1+q^{2n}}\cos(\pi n u/K) $$ where $$K=K(k) =\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}},k'=\sqrt{1-k^2},K'=K(k'),q=e^{-\pi K'/K} $$ Expanding $\cos(\pi nu/K) $ as a power series in $u$ and equating coefficients of $u^2$ in the two series for $\operatorname {dn} (u, k) $ we get $$k^2=\frac{2\pi^3}{K^3}\sum_{n=1}^{\infty} \frac{n^2q^n}{1+q^{2n}}$$ or $$g(q) =\sum_{n=1}^{\infty} \frac{n^2q^n}{1+q^{2n}}=\frac{k^2K^3}{2\pi^3}$$ Let $l, l', L, L'$ correspond to $q^2$ in the same manner as $k, k', K, K'$ correspond to $q$ then we have $$g(q^2)=\frac{l^2L^3}{2\pi^3}$$ By Landen transformation we have $$k=\frac{2\sqrt{l}}{1+l},K=(1+l)L$$ and then we get $$g(q) =\frac{4l(1+l)L^3}{2\pi^3}$$ Thus we have $$2g(q)-8g(q^2)=\frac{4lL^3}{\pi^3}\tag{4}$$ For $q^2=e^{-\pi} $ we have $$l=1/\sqrt{2},L=\Gamma^2(1/4)/(4\sqrt{\pi})$$ and thus $$B=\frac{\sqrt{2}\Gamma ^6(1/4)}{32\pi^{9/2}}-\frac{2q}{1+q^2}$$ The desired sum is $A-B$ where $A$ has been obtained in equation $(3)$.
