If b = a - frac{a^2}{2} + frac{a^3}{3} - frac | vedclass

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(C) Given the logarithmic series expansion: $b = a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \dots = \log_e(1 + a)$.This implies $1 + a = e^b$

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$a$

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(C) Given the logarithmic series expansion: $b = a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \dots = \log_e(1 + a)$.This implies $1 + a = e^b$.Using the exponential series expansion for $e^b$:$e^b = 1 + \frac{b}{1!} + \frac{b^2}{2!} + \frac{b^3}{3!} + \dots$Substituting $1 + a$ for $e^b$:$1 + a = 1 + b + \frac{b^2}{2!} + \frac{b^3}{3!} + \dots$Subtracting $1$ from both sides:$a = b + \frac{b^2}{2!} + \frac{b^3}{3!} + \dots$Thus,the value is $a$.

If $b = a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \dots$,then $b + \frac{b^2}{2!} + \frac{b^3}{3!} + \frac{b^4}{4!} + \dots \infty = $

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