The sum of the series log_{4} 2 - log_{8} 2 + | vedclass

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

Question

(D) Given the series $S = \log_{4} 2 - \log_{8} 2 + \log_{16} 2 - \dots$Using the property $\log_{y^n} x^m = \frac{m}{n} \log_{y} x$,we have $\log_{2^

Vedclass · Accepted Answer

$1 - \log_{e} 2$

Vedclass · Answer

(D) Given the series $S = \log_{4} 2 - \log_{8} 2 + \log_{16} 2 - \dots$Using the property $\log_{y^n} x^m = \frac{m}{n} \log_{y} x$,we have $\log_{2^2} 2 = \frac{1}{2} \log_{2} 2 = \frac{1}{2}$,$\log_{2^3} 2 = \frac{1}{3}$,and so on.Thus,$S = \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \dots$We know the logarithmic expansion $\log_{e}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$For $x=1$,$\log_{e}(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$Rearranging this,$\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots = 1 - \log_{e} 2$.Therefore,$S = 1 - \log_{e} 2$.

The sum of the series $\log_{4} 2 - \log_{8} 2 + \log_{16} 2 - \dots$ is

Similar Questions

The sum of the infinite series $\frac{1}{1 \times 2} - \frac{1}{2 \times 3} + \frac{1}{3 \times 4} - \dots \infty$ is equal to:

$\frac{1}{3} + \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} + \frac{1}{4 \cdot 3^4} + \dots \infty = $

If $0 < y < 2^{1/3}$ and $x(y^3 - 1) = 1$,then $\frac{2}{x} + \frac{2}{3x^3} + \frac{2}{5x^5} + \dots$ is equal to:

If $|x| < 1$,then the coefficient of $x^5$ in the expansion of $(1 - x) \ln(1 - x)$ is

$\cosh^{-1} 2 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module