frac{1}{2}x^2 + frac{2}{3}x^3 + frac{3}{4}x^4 | vedclass

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(D) ધારો કે $S = \sum_{n=1}^{\infty} \frac{n}{n+1} x^{n+1}$.આપણે $\frac{n}{n+1} = 1 - \frac{1}{n+1}$ લખી શકીએ.તેથી,$S = \sum_{n=1}^{\infty} (1 - \frac

Vedclass · Accepted Answer

$\frac{x}{1 - x} + \log_e(1 - x)$

Vedclass · Answer

(D) ધારો કે $S = \sum_{n=1}^{\infty} \frac{n}{n+1} x^{n+1}$.આપણે $\frac{n}{n+1} = 1 - \frac{1}{n+1}$ લખી શકીએ.તેથી,$S = \sum_{n=1}^{\infty} (1 - \frac{1}{n+1}) x^{n+1} = \sum_{n=1}^{\infty} x^{n+1} - \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$.$S = (x^2 + x^3 + x^4 + \dots) - (\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots)$.પ્રથમ ભાગ એક ભૌમિતિક શ્રેણી છે: $\frac{x^2}{1-x}$.બીજો ભાગ $-\log_e(1-x) - x$ નું વિસ્તરણ છે.તેથી,$S = \frac{x^2}{1-x} - (-\log_e(1-x) - x) = \frac{x^2}{1-x} + x + \log_e(1-x)$.$S = \frac{x^2 + x(1-x)}{1-x} + \log_e(1-x) = \frac{x}{1-x} + \log_e(1-x)$.

$\frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 + \dots \infty = $

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