Question

Evaluate:${\sum }_{m=1}^{\infty }{\sum }_{n=1}^{\infty }\frac{m^{2}n}{3^m(n{.3}^m+m{.3}^n)}$

• A. $\frac{9}{16}$
• B. $\frac{9}{32}$
• C. $\frac{9}{8}$
• D. $\frac{3}{4}$

Answer 1

Answer: (B)

$\frac{9}{32}$

Let $S={\sum }_{m=1}^{\infty }{\sum }_{n=1}^{\infty }\frac{m^{2}n}{3^m(n{.3}^m+m{.3}^n)}$
$={\sum }_{m=1}^{\infty }{\sum }_{n=1}^{\infty }\frac{1}{\left(\frac{3^m}{m}\right)\left(\frac{3^m}{m}+\frac{3^n}{n}\right)}$
$S={\sum }_{m=1}^{\infty }{\sum }_{n=1}^{\infty }\frac{1}{a_m(a_m+a_n)}\cdots \left(1\right)$
(Let $a_m=\frac{3^m}{m}and{a}_n=\frac{3^n}{n}$)
By interchanging $m$ and $n$, then
$S={\sum }_{m=1}^{\infty }{\sum }_{n=1}^{\infty }\frac{1}{a_n(a_n+a_m)}\cdots \left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$, then
$2S={\sum }_{m=1}^{\infty }{\sum }_{n=1}^{\infty }\frac{1}{a_m{a}_n}$
$={\sum }_{m=1}^{\infty }{\sum }_{n=1}^{\infty }\frac{mn}{3^{m+n}}={\left(\sum _{n=1}^{\infty }\frac{n}{3^n}\right)}^2$
$={\left(\frac{3}{4}\right)}^2$
$=\frac{9}{16}$
$\Rightarrow S=\frac{9}{32}$
Hence, option 'B' is correct.