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Evaluate \sum _{{m=1}}^{\infty }\sum _{{n=1}}^{\infty }{\frac  {m^{2}\,n}{3^{m}\left(m\,3^{n}+n\,3^{m}\right)}}.

Solution: First rewrite the summand as {\frac  {m}{3^{m}}}{\frac  {m\,n}{m\,3^{n}+n\,3^{m}}}.

Then make the substitution a_{j}={\frac  {3^{j}}{j}}.

Now call the sum S, for notational purposes.

With this notation it follows S=\sum _{{m=1}}^{\infty }\sum _{{n=1}}^{\infty }{\frac  {1}{a_{m}}}{\frac  {1}{a_{m}+a_{n}}}.

Then recognize S=\sum _{{m=1}}^{\infty }\sum _{{n=1}}^{\infty }{\frac  {1}{a_{n}}}{\frac  {1}{a_{m}+a_{n}}}. Only here the indices have been reversed, first in name and then in order.

Adding the two previous expressions for S gives 2S=\sum _{{m=1}}^{\infty }\sum _{{n=1}}^{\infty }\left({\frac  {1}{a_{m}}}+{\frac  {1}{a_{n}}}\right)\left({\frac  {1}{a_{m}+a_{n}}}\right).

But {\frac  {1}{a_{m}}}+{\frac  {1}{a_{n}}}={\frac  {a_{m}+a_{n}}{a_{m}\,a_{n}}}. So it follows that 2S=\sum _{{m=1}}^{\infty }\sum _{{n=1}}^{\infty }{\frac  {1}{a_{m}\,a_{n}}} which is equivalent to \left(\sum _{{m=1}}^{\infty }{\frac  {1}{a_{m}}}\right)^{2}.

So now all that is left is to evaluate \sum _{{m=1}}^{\infty }{\frac  {1}{a_{m}}}=\sum _{{m=1}}^{\infty }{\frac  {m}{3^{m}}}, and call this A\,. Now it follows from A=\sum _{{m=1}}^{\infty }{\frac  {m}{3^{m}}}={\frac  {1}{3}}+{\frac  {2}{9}}+{\frac  {3}{27}}+... that 3A=1+{\frac  {2}{3}}+{\frac  {3}{9}}+....

Subtracting gives 2A=1+{\frac  {1}{3}}+{\frac  {1}{9}}+...={\frac  {1}{1-1/3}}={\frac  {3}{2}}. So A={\frac  {3}{4}}, and since 2S=A^{2}\, it follows that S={\frac  {9}{32}}.


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