Q. If $|a| < 1,\ b = \sum_{k = 1}^{\infty}\frac{a^{k}}{k}$, then $a$ is equal to
A) $\sum_{k = 1}^{\infty}\frac{( - 1)^{k}b^{k}}{k}$
B) $\sum_{k = 1}^{\infty}\frac{( - 1)^{k - 1}b^{k}}{k!}$
C) $\sum_{k = 1}^{\infty}\frac{( - 1)^{k}b^{k}}{(k - 1)!}$
D) $\sum_{k = 1}^{\infty}\frac{( - 1)^{k - 1}b^{k}}{(k + 1)!}$
Explanation
(b)
Given, $b = \frac{a^{1}}{1} + \frac{a^{2}}{2} + \frac{a^{3}}{3} + ...\infty$
$\Rightarrow \ \ b = - \log(1 - a)\ \ \ \lbrack\because|a| < 1\rbrack$
$\Rightarrow \ \ e^{- b} = (1 - a)$
$\Rightarrow \ \ \ a = \frac{b}{1!} - \frac{b^{2}}{2!} + \frac{b^{3}}{3!} - ...\infty$
$= \sum_{k = 1}^{\infty}\frac{( - 1)^{k - 1}b^{k}}{k!}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.