Question

If$\left|a\right|<1,b=\sum _{n=1}^{\infty }\frac{a^k}{k}$ then $a$ is equal to

• A. $\left(1\right)\sum _{k=1}^{\infty }{(-1)}^k\frac{b^k}{k}$
• B. $\left(1\right)\sum _{k=1}^{\infty }{(-1)}^{k-1}\frac{b^k}{k!}$
• C. $\left(1\right)\sum _{k=1}^{\infty }{(-1)}^{k-1}\frac{b^k}{(k-1)!}$
• D. None of these

Answer 1

The correct option is B

$\left(1\right)\sum _{k=1}^{\infty }{(-1)}^{k-1}\frac{b^k}{k!}$

Explanation for the correct option:

Find the value of $\mathit{a}$:

Given $\left|a\right|<1,b=\sum _{n=1}^{\infty }\frac{a^k}{k}$

$$\begin{gathered} =\frac{a^1}{1}+\frac{a^2}{2}+\frac{a^3}{3}+...........\infty \\ =-\log (1-a)\mathbf{\left[}\mathbf{\because }\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{x}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\infty }\mathbf{)}\mathbf{\right]} \end{gathered}$$

$\begin{array}{rcl}& \Rightarrow & e^{-b}=(1-a)\\ & \Rightarrow & a=1-e^{-b}\\ & \Rightarrow & a=1-\left[1-b+\frac{b^2}{2!}-\frac{b^3}{3!}+........\infty \right]\\ & \Rightarrow & a=b-\frac{b^2}{2!}+\frac{b^3}{3!}+........\infty \\ & \Rightarrow & a=\sum _{k=1}^{\infty }{(-1)}^{k-1}\frac{b^k}{k!}\end{array}$

Hence, the correct option is B.