Question

$\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }\sum _{k=1}^{\infty }\frac{1}{a^{i+j+k}}$ is equal to $(\text{where}|a|>1)$

• A. $\frac{3}{a-1}$
• B. None of these
• C. $\frac{3}{a^3-1}$
• D. $(a-1{)}^{-3}$

Answer 1

The correct option is D $(a-1{)}^{-3}$

$\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }\sum _{k=1}^{\infty }\frac{1}{a^{i+j+k}}$
$=\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }\sum _{k=1}^{\infty }\frac{1}{a^i.a^j.a^k}$

$=\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }\frac{1}{a^i.a^j}.\left(\frac{\frac{1}{a}}{1-\frac{1}{a}}\right)$

$=\frac{1}{a-1}\sum _{i=1}^{\infty }\sum _{j=1}^{\infty }\frac{1}{a^i\cdot a^j}$
$=\frac{1}{a-1}\sum _{i=1}^{\infty }\frac{1}{a^i}\times \left(\frac{\frac{1}{a}}{1-\frac{1}{a}}\right)$

$=\frac{1}{(a-1{)}^2}\sum _{i=1}^{\infty }\frac{1}{a^i}=\frac{1}{(a-1{)}^3}=(a-1{)}^{-3}$