frac{x - 1}{x + 1} + frac{1}{2} cdot frac{x^2 | vedclass

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(A) ધારો કે $S = \sum_{n=1}^{\infty} \frac{1}{n} \left( \frac{x^n - 1}{(x + 1)^n} \right) = \sum_{n=1}^{\infty} \frac{1}{n} \left( \left( \frac{x}{x +

Vedclass · Accepted Answer

$\log_e x$

Vedclass · Answer

(A) ધારો કે $S = \sum_{n=1}^{\infty} \frac{1}{n} \left( \frac{x^n - 1}{(x + 1)^n} \right) = \sum_{n=1}^{\infty} \frac{1}{n} \left( \left( \frac{x}{x + 1} \right)^n - \left( \frac{1}{x + 1} \right)^n \right)$.લોગરીધમિક શ્રેણીના વિસ્તરણ $\ln(1 - y) = - \sum_{n=1}^{\infty} \frac{y^n}{n}$ નો ઉપયોગ કરતા:$S = - \ln\left(1 - \frac{x}{x + 1}\right) - \left( - \ln\left(1 - \frac{1}{x + 1}\right) \right)$.$S = - \ln\left(\frac{1}{x + 1}\right) + \ln\left(\frac{x}{x + 1}\right)$.$S = \ln(x + 1) + \ln\left(\frac{x}{x + 1}\right) = \ln\left((x + 1) \cdot \frac{x}{x + 1}\right) = \ln x$.

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

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