To prove the identity $$4\sum_{m,n=1}^{\infty}\frac{q^{n+m}}{(1+q^n)(1+q^m)}(z^{n-m}+z^{m-n})=8\sum_{m,n=1}^{\infty}\frac{q^{n+2m}}{(1+q^n)(1+q^{n+m})}(z^m+z^{-m})$$ I replaced $m-n$ by $k$ in LHS and got $$4\sum_{n=1}^{\infty}\sum_{k=-\infty}^{\infty}\frac{q^{2n+k}}{(1+q^n)(1+q^{n+k})}(z^{-k}+z^k)\\=4\sum_{n=1}^{\infty}\Big(\sum_{k=-\infty}^{0}+\sum_{k=1}^{\infty}\Big)\frac{q^{2n+k}}{(1+q^n)(1+q^{n+k})}(z^{-k}+z^k) $$if we can show that $$\sum_{n=1}^{\infty}\sum_{k=-\infty}^{0}\frac{q^{2n+k}}{(1+q^n)(1+q^{n+k})}(z^{-k}+z^k)=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{q^{2n+k}}{(1+q^n)(1+q^{n+k})}(z^{-k}+z^k)$$ then we are done. In this direction, I proceeded as replacing $k $ by $-k$ in LHS we get, $$\sum_{n=1}^{\infty}\sum_{k=0}^{\infty}\frac{q^{2n-k}}{(1+q^n)(1+q^{n-k})}(z^k+z^{-k})\\=2\sum_{n=1}^{\infty}\frac{q^{2n}}{(1+q^n)^2}+\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{q^{2n-k}}{(1+q^n)(1+q^{n-k})}(z^k+z^{-k})$$ How to go ahead?
$4\sum_{m,n=1}^{\infty}\frac{q^{n+m}}{(1+q^n)(1+q^m)}(z^{n-m}+z^{m-n})=8\sum_{m,n=1}^{\infty}\frac{q^{n+2m}}{(1+q^n)(1+q^{n+m})}(z^m+z^{-m})$
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sequences-and-series
elementary-number-theory
integer-partitions
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0$$ \begin{align} \,& q=1/4,\,z=1/2\space\implies \\ \,& 4\,\sum_{m=1}^{\infty}\,\sum_{n=1}^{\infty}\,\frac{q^{m}}{1+q^{m}}\,\frac{q^{n}}{1+q^{n}}\,\left(z^{m-n}+z^{n-m}\right)\approx\color{red}{0.826} \\ \,& 8\,\sum_{m=1}^{\infty}\,\sum_{n=1}^{\infty}\,\frac{q^{m}}{1+q^{n}}\,\frac{q^{n+m}}{1+q^{n+m}}\,\left(z^{+m}+z^{-m}\right)\approx\color{red}{0.375} \end{align} $$ – 2017-02-05