frac{1}{n^2} + frac{1}{2n^4} + frac{1}{3n^6} | vedclass

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(C) The given series is $S = \frac{1}{n^2} + \frac{1}{2n^4} + \frac{1}{3n^6} + \dots \infty$. This can be written as $S = \frac{(1/n^2)^1}{1} + \frac{

Vedclass · Accepted Answer

$\log_e \left( \frac{n^2}{n^2 - 1} \right)$

Vedclass · Answer

(C) The given series is $S = \frac{1}{n^2} + \frac{1}{2n^4} + \frac{1}{3n^6} + \dots \infty$. This can be written as $S = \frac{(1/n^2)^1}{1} + \frac{(1/n^2)^2}{2} + \frac{(1/n^2)^3}{3} + \dots$. Using the logarithmic expansion $-\log_e(1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dots$ for $|x| $S = -\log_e(1 - \frac{1}{n^2}) = -\log_e(\frac{n^2 - 1}{n^2}) = \log_e(\frac{n^2}{n^2 - 1})$. Thus,the correct option is $C$.

$\frac{1}{n^2} + \frac{1}{2n^4} + \frac{1}{3n^6} + \dots \infty = $

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