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

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

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

Vedclass · Accepted Answer

$1 - \ln 2$

Vedclass · Answer

(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:

Similar Questions

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

For $|x| < 1$,the coefficient of $x^3$ in the expansion of $\log(1+x+x^2)$ in ascending powers of $x$ is:

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

$1 + \frac{2}{3} - \frac{2}{4} + \frac{2}{5} - \dots \infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module