The value of the infinite series log _4 2 - l | vedclass

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Question

(A) The given series is $S = \log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$. Using the property $\log _{a^n} b = \frac{1}{n} \log _a b$,we can w

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$1 - \ln 2$

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(A) The given series is $S = \log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$. Using the property $\log _{a^n} b = \frac{1}{n} \log _a b$,we can write: $\log _4 2 = \log _{2^2} 2 = \frac{1}{2} \log _2 2 = \frac{1}{2}$ $\log _8 2 = \log _{2^3} 2 = \frac{1}{3} \log _2 2 = \frac{1}{3}$ $\log _{16} 2 = \log _{2^4} 2 = \frac{1}{4} \log _2 2 = \frac{1}{4}$ Substituting these into the series: $S = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$ Recall the Taylor series expansion for $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ For $x=1$,$\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ Rearranging this,we get $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \ln 2$. Thus,the sum of the series is $1 - \ln 2$.

The value of the infinite series $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$ is:

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