# Calc1.11

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 $S={\frac {9}{32}}$.