The sum to infinity of the given series frac{ | vedclass

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(A) The given series is $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ This can be rewritten as $\frac{1}{n} - \frac{(1/n)^2

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

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(A) The given series is $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ This can be rewritten as $\frac{1}{n} - \frac{(1/n)^2}{2} + \frac{(1/n)^3}{3} - \frac{(1/n)^4}{4} + \dots$ We know the logarithmic series expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ for $|x| Substituting $x = \frac{1}{n}$,we get $\log_e(1 + \frac{1}{n}) = \frac{1}{n} - \frac{(1/n)^2}{2} + \frac{(1/n)^3}{3} - \dots$ Therefore,the sum is $\log_e\left(\frac{n+1}{n}\right)$.

The sum to infinity of the given series $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ is

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