frac{1}{x + 1} + frac{1}{2(x + 1)^2} + frac{1 | vedclass

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

Question

(A) The given series is $\frac{1}{x + 1} + \frac{1}{2(x + 1)^2} + \frac{1}{3(x + 1)^3} + \dots \infty$. We know the logarithmic expansion: $-\log_e(1

Vedclass · Accepted Answer

$\log_e\left(1 + \frac{1}{x}\right)$

Vedclass · Answer

(A) The given series is $\frac{1}{x + 1} + \frac{1}{2(x + 1)^2} + \frac{1}{3(x + 1)^3} + \dots \infty$. We know the logarithmic expansion: $-\log_e(1 - y) = y + \frac{y^2}{2} + \frac{y^3}{3} + \dots \infty$ for $|y| Let $y = \frac{1}{x + 1}$. Then the series becomes $-\log_e\left(1 - \frac{1}{x + 1}\right)$. $= -\log_e\left(\frac{x + 1 - 1}{x + 1}\right) = -\log_e\left(\frac{x}{x + 1}\right)$. $= \log_e\left(\frac{x + 1}{x}\right) = \log_e\left(1 + \frac{1}{x}\right)$.

$\frac{1}{x + 1} + \frac{1}{2(x + 1)^2} + \frac{1}{3(x + 1)^3} + \dots \infty = $

Similar Questions

The expression $\log_{e} 2 + \log_{e} \left( 1 + \frac{1}{2} \right) + \log_{e} \left( 1 + \frac{1}{3} \right) + \dots + \log_{e} \left( 1 + \frac{1}{n - 1} \right)$ is equal to

$e^{\left( {x - \frac{1}{2}{(x - 1)}^2 + \frac{1}{3}{(x - 1)}^3 - \frac{1}{4}{(x - 1)}^4 + \dots} \right)}$ is equal to

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

$\log _4 2 - \log _8 2 + \log _{16} 2 - \ldots$ is equal to

$\frac{1}{1 \cdot 3} + \frac{1}{2} \cdot \frac{1}{3 \cdot 5} + \frac{1}{3} \cdot \frac{1}{5 \cdot 7} + \dots \infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module