frac{x - 1}{x + 1} + frac{1}{2} cdot frac{x^2 | vedclass

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

Question

(A) माना $S = \sum_{n=1}^{\infty} \frac{1}{n} \left( \frac{x^n - 1}{(x + 1)^n} \right) = \sum_{n=1}^{\infty} \frac{1}{n} \left( \left( \frac{x}{x + 1}

Vedclass · Accepted Answer

$\log_e x$

Vedclass · Answer

(A) माना $S = \sum_{n=1}^{\infty} \frac{1}{n} \left( \frac{x^n - 1}{(x + 1)^n} \right) = \sum_{n=1}^{\infty} \frac{1}{n} \left( \left( \frac{x}{x + 1} \right)^n - \left( \frac{1}{x + 1} \right)^n \right)$.लघुगणकीय श्रेणी विस्तार $\ln(1 - y) = - \sum_{n=1}^{\infty} \frac{y^n}{n}$ का उपयोग करते हुए:$S = - \ln\left(1 - \frac{x}{x + 1}\right) - \left( - \ln\left(1 - \frac{1}{x + 1}\right) \right)$.$S = - \ln\left(\frac{1}{x + 1}\right) + \ln\left(\frac{x}{x + 1}\right)$.$S = \ln(x + 1) + \ln\left(\frac{x}{x + 1}\right) = \ln\left((x + 1) \cdot \frac{x}{x + 1}\right) = \ln x$.

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

Similar Questions

$\frac{1}{2} + \frac{3}{2} \cdot \frac{1}{4} + \frac{5}{3} \cdot \frac{1}{8} + \frac{7}{4} \cdot \frac{1}{16} + \dots \infty = $

श्रेणी $\log_{4} 2 - \log_{8} 2 + \log_{16} 2 - \dots$ का योग क्या है?

अनंत श्रेणी $\log _4 2 - \log _8 2 + \log _{16} 2 - \dots \infty$ का मान है:

$\left( \frac{a - b}{a} \right) + \frac{1}{2} \left( \frac{a - b}{a} \right)^2 + \frac{1}{3} \left( \frac{a - b}{a} \right)^3 + \dots = $

$\log_e \left( 1 + ax^2 + a^2 + \frac{a}{x^2} \right)$ का मान क्या है?

Vedclass Test Series

Exam Paper Generator

Online Exam Module