If |a|

Question

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

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

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(B) Given that $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$.We know the logarithmic series expansion: $-\ln(1-a) = \sum_{k=1}^{\infty} \frac{a^k}{k}$ for $|a| Therefore,$b = -\ln(1-a)$.This implies $e^{-b} = 1-a$,so $a = 1 - e^{-b}$.Using the Taylor series expansion for $e^{-b} = \sum_{k=0}^{\infty} \frac{(-b)^k}{k!} = 1 - b + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots$.Substituting this into the expression for $a$:$a = 1 - (1 - b + \frac{b^2}{2!} - \frac{b^3}{3!} + \dots) = b - \frac{b^2}{2!} + \frac{b^3}{3!} - \dots$.This can be written as $a = \sum_{k=1}^{\infty} \frac{(-1)^{k-1} b^k}{k!}$.

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

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