# Simplification of double series

Consider a double series in the following form: $$\sum_{j,k=0}^\infty\frac{1}{a_jb_k+c_jd_k}.$$ Is there a way of evaluating it, if $\sum_{j=0}^\infty a_j^{-1}$, $\sum_{k=0}^\infty b_k^{-1}$, etc... are all known?

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One does not "solve" a series. Do you mean, "evaluate" it? – Gerry Myerson Feb 11 '13 at 23:07
Yes, you are right, I mean evaluate it! – Ziofil Feb 18 '13 at 17:07
Good. Then, why not edit the question, so it says what you mean? – Gerry Myerson Feb 18 '13 at 22:57

Let $a_j=b_j=c_j=j^2$, $d_j=\alpha j$, and make all the sums start at $1$ instead of zero. Then the double sum is $$\sum_{j,k}{1\over j^2k^2+\alpha j^2k}=\left(\sum_j{1\over j^2}\right)\left(\sum_k{1\over k^2+\alpha k}\right)$$ which clearly depends on $\alpha$. But $\sum_k d_k^{-1}$ diverges, regardless of the value of $\alpha$.
I'm sure there are examples of sequences $a_j$ and $a_j'$ which lead to different values of the double sum even though $\sum a_j$ and $\sum a_j'$ both converge and are equal, but I don't have the imagination to construct one where it easy to compute the double sums to see the difference.