frac{1}{2}x^2 + frac{2}{3}x^3 + frac{3}{4}x^4 | vedclass

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

Question

(D) Let $S = \sum_{n=1}^{\infty} \frac{n}{n+1} x^{n+1}$.We can write $\frac{n}{n+1} = 1 - \frac{1}{n+1}$.So,$S = \sum_{n=1}^{\infty} (1 - \frac{1}{n+1

Vedclass · Accepted Answer

$\frac{x}{1 - x} + \log_e(1 - x)$

Vedclass · Answer

(D) Let $S = \sum_{n=1}^{\infty} \frac{n}{n+1} x^{n+1}$.We can write $\frac{n}{n+1} = 1 - \frac{1}{n+1}$.So,$S = \sum_{n=1}^{\infty} (1 - \frac{1}{n+1}) x^{n+1} = \sum_{n=1}^{\infty} x^{n+1} - \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}$.$S = (x^2 + x^3 + x^4 + \dots) - (\frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dots)$.The first part is a geometric series: $\frac{x^2}{1-x}$.The second part is the expansion of $-\log_e(1-x) - x$.Therefore,$S = \frac{x^2}{1-x} - (-\log_e(1-x) - x) = \frac{x^2}{1-x} + x + \log_e(1-x)$.$S = \frac{x^2 + x(1-x)}{1-x} + \log_e(1-x) = \frac{x}{1-x} + \log_e(1-x)$.

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

Similar Questions

$\log_e [(1 + x)^{1 + x} (1 - x)^{1 - x}] = $

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:

$1 + \left( \frac{1}{2} + \frac{1}{3} \right) \frac{1}{4} + \left( \frac{1}{4} + \frac{1}{5} \right) \frac{1}{4^2} + \left( \frac{1}{6} + \frac{1}{7} \right) \frac{1}{4^3} + \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

In the expansion of $2 \ln x - \ln(x + 1) - \ln(x - 1)$,the coefficient of $x^{-4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module