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If $|a| < 1, b=\sum_{k=1}^{\infty} \frac{a^k}{k}$, then $a$ is equal to
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Verified Answer
The correct answer is:
$\sum_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k !}$
Given that
$\begin{aligned}
b & =\sum_{k-1}^{\infty} \frac{a^k}{k} \\
& =\frac{a^1}{1}+\frac{a^2}{2}+\frac{a^3}{3}+\ldots \infty \\
\Rightarrow \quad b & =-\log (1-a) \quad \quad[\because|a| < 1] \\
\Rightarrow \quad e^{-b} & =(1-a) \\
\Rightarrow a=1-e^{-b} & =1-\left(1-\frac{b}{1 !}+\frac{b^2}{2 !}-\frac{b^3}{3 !}+\ldots \infty\right) \\
\Rightarrow \quad a & =\frac{b}{1 !}-\frac{b^2}{2 !}+\frac{b^3}{3 !}-\ldots \infty \\
\Rightarrow \quad a & =\sum_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k !}
\end{aligned}$
$\begin{aligned}
b & =\sum_{k-1}^{\infty} \frac{a^k}{k} \\
& =\frac{a^1}{1}+\frac{a^2}{2}+\frac{a^3}{3}+\ldots \infty \\
\Rightarrow \quad b & =-\log (1-a) \quad \quad[\because|a| < 1] \\
\Rightarrow \quad e^{-b} & =(1-a) \\
\Rightarrow a=1-e^{-b} & =1-\left(1-\frac{b}{1 !}+\frac{b^2}{2 !}-\frac{b^3}{3 !}+\ldots \infty\right) \\
\Rightarrow \quad a & =\frac{b}{1 !}-\frac{b^2}{2 !}+\frac{b^3}{3 !}-\ldots \infty \\
\Rightarrow \quad a & =\sum_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k !}
\end{aligned}$
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