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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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)

Vedclass · Accepted Answer

$P = 2Q$

Vedclass · Answer

(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

Similar Questions

$1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \dots = $

If $|a| < 1$ and $b = \sum_{k=1}^{\infty} \frac{a^k}{k}$,then $a$ is equal to

$1+\frac{1}{3 \cdot 2^2}+\frac{1}{5 \cdot 2^4}+\frac{1}{7 \cdot 2^6}+\ldots$ is equal to

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

$\frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 5} + \frac{1}{6 \cdot 7} + \frac{1}{8 \cdot 9} + \dots$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module