Let \begin{align} L&=2^{3 / 2} \pi^{4} \frac{1}{4} \sum_{n, m, k=1}^{\infty} n^{2} m a_{n} a_{m} b_{k}\left(\delta_{m, n+k}+\delta_{n, m+k}-\delta_{k, n+m}\right) \\[5pt] R_{1}&=2^{3 / 2} \pi^{4} \frac{1}{4} \sum_{n, m, k=1}^{\infty} n^{3 / 2} m^{3 / 2} a_{n} a_{m} b_{k}\left(\delta_{m, n+k}+\delta_{n, m+k}+\delta_{k, n+m}\right) \\[5pt] R_{2}&=\frac{1}{2} 2^{3 / 2} \pi^{4} \frac{1}{4} \sum_{n, m, k=1}^{\infty} n^{1 / 2} m^{1 / 2} k^{2} a_{n} a_{m} b_{k}\left(\delta_{m, n+k}+\delta_{n, m+k}-\delta_{k, n+m}\right), \end{align} where the $\delta$ denote the Kronecker delta.
How can we compute $$L-R_1-R_2 \ ?$$
The motivation for this computation comes from an integration by parts formula analyzed on MathOverflow.
With the help of the Kronecker delta's we can carry out the summation over $k$, $$L-R_1-R_2=2^{3/2}\pi^4 \frac{1}{4}\sum_{n=1}^\infty\sum_{m=1}^\infty a_na_m\biggl[\bigl(m n^2-(m n)^{3/2}-(m-n)^2 \sqrt{m n}\bigr) b_{|n-m|}$$ $$-\bigl(m n^2+(m n)^{3/2}-(m+n)^2 \sqrt{m n}\bigr) b_{n+m}\biggr].$$ Without further knowledge of the coefficients this cannot be evaluated further.