frac{1}{x + 1} + frac{1}{2(x + 1)^2} + frac{1 | vedclass

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Question

(A) The given series is $\frac{1}{x + 1} + \frac{1}{2(x + 1)^2} + \frac{1}{3(x + 1)^3} + \dots \infty$. We know the logarithmic expansion: $-\log_e(1

Vedclass · Accepted Answer

$\log_e\left(1 + \frac{1}{x}\right)$

Vedclass · Answer

(A) The given series is $\frac{1}{x + 1} + \frac{1}{2(x + 1)^2} + \frac{1}{3(x + 1)^3} + \dots \infty$. We know the logarithmic expansion: $-\log_e(1 - y) = y + \frac{y^2}{2} + \frac{y^3}{3} + \dots \infty$ for $|y| Let $y = \frac{1}{x + 1}$. Then the series becomes $-\log_e\left(1 - \frac{1}{x + 1}\right)$. $= -\log_e\left(\frac{x + 1 - 1}{x + 1}\right) = -\log_e\left(\frac{x}{x + 1}\right)$. $= \log_e\left(\frac{x + 1}{x}\right) = \log_e\left(1 + \frac{1}{x}\right)$.

$\frac{1}{x + 1} + \frac{1}{2(x + 1)^2} + \frac{1}{3(x + 1)^3} + \dots \infty = $

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