If |a|

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(B) Given that $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$.Using the logarithmic series expansion,we know that $-\ln(1-a) = \sum_{k=1}^{\infty} \frac{a^k}

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$\sum_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k!}$

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(B) Given that $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$.Using the logarithmic series expansion,we know that $-\ln(1-a) = \sum_{k=1}^{\infty} \frac{a^k}{k}$ for $|a| Therefore,$b = -\ln(1-a)$.This implies $e^{-b} = 1-a$,so $a = 1 - e^{-b}$.Using the Taylor series expansion for $e^{-b} = 1 - \frac{b}{1!} + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots$,we get:$a = 1 - (1 - \frac{b}{1!} + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots)$$a = \frac{b}{1!} - \frac{b^2}{2!} + \frac{b^3}{3!} - \dots$$a = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k!}$.

If $|a| < 1$ and $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$,then $a$ is equal to

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