If P = 1 + frac{1}{2 times 2} + frac{1}{3 tim | vedclass

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(C) Given, $P = 1 + \frac{1}{2 	imes 2} + \frac{1}{3 	imes 2^{2}} + \dots = \sum_{n=1}^{\infty} \frac{(1/2)^{n-1}}{n}$.Using the expansion $-\ln(1-x

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$P = 2Q$

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(C) Given, $P = 1 + \frac{1}{2 	imes 2} + \frac{1}{3 	imes 2^{2}} + \dots = \sum_{n=1}^{\infty} \frac{(1/2)^{n-1}}{n}$.Using the expansion $-\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$, we have $P = 2 \sum_{n=1}^{\infty} \frac{(1/2)^n}{n} = 2 [-\ln(1 - 1/2)] = 2 \ln 2$.Given, $Q = \frac{1}{1 	imes 2} + \frac{1}{3 	imes 4} + \frac{1}{5 	imes 6} + \dots = \sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n)}$.Using partial fractions, $\frac{1}{(2n-1)(2n)} = \frac{1}{2n-1} - \frac{1}{2n}$.Thus, $Q = (1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + \dots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} = \ln(1+1) = \ln 2$.Comparing $P = 2 \ln 2$ and $Q = \ln 2$, we get $P = 2Q$.

If $P = 1 + \frac{1}{2 \times 2} + \frac{1}{3 \times 2^{2}} + \dots$ and $Q = \frac{1}{1 \times 2} + \frac{1}{3 \times 4} + \frac{1}{5 \times 6} + \dots$, then

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