frac{1}{1 cdot 3} + frac{1}{2} cdot frac{1}{3 | vedclass

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(A) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)}$.Using partial fractions,$\frac{1}{(2n-1)(2n+1)} = \frac{1}{2} \left( \frac{1

Vedclass · Accepted Answer

$2 \log_e 2 - 1$

Vedclass · Answer

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)}$.Using partial fractions,$\frac{1}{(2n-1)(2n+1)} = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right)$.So,$S = \sum_{n=1}^{\infty} \frac{1}{2n} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) = \sum_{n=1}^{\infty} \left( \frac{1}{2n(2n-1)} - \frac{1}{2n(2n+1)} \right)$.This can be written as $S = \sum_{n=1}^{\infty} \left( \left( \frac{1}{2n-1} - \frac{1}{2n} \right) - \left( \frac{1}{2n} - \frac{1}{2n+1} \right) \right)$.Expanding the series: $S = (1 - 1/2) - (1/2 - 1/3) + (1/3 - 1/4) - (1/4 - 1/5) + \dots$$S = 1 - 2(1/2) + 2(1/3) - 2(1/4) + 2(1/5) - \dots$$S = -1 + 2(1 - 1/2 + 1/3 - 1/4 + 1/5 - \dots) = -1 + 2 \log_e 2$.

$\frac{1}{1 \cdot 3} + \frac{1}{2} \cdot \frac{1}{3 \cdot 5} + \frac{1}{3} \cdot \frac{1}{5 \cdot 7} + \dots \infty = $

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