The coefficient of n^{-r} in the expansion of | vedclass

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

Question

(A) We have $\log_{10}\left(\frac{n}{n-1}\right) = \log_e\left(\frac{n}{n-1}\right) \cdot \log_{10}e$. $= -\log_e\left(\frac{n-1}{n}\right) \cdot \log

Vedclass · Accepted Answer

$\frac{1}{r \log_e 10}$

Vedclass · Answer

(A) We have $\log_{10}\left(\frac{n}{n-1}\right) = \log_e\left(\frac{n}{n-1}\right) \cdot \log_{10}e$. $= -\log_e\left(\frac{n-1}{n}\right) \cdot \log_{10}e = -\log_{10}e \cdot \log_e\left(1 - \frac{1}{n}\right)$. Using the logarithmic series expansion $\log_e(1-x) = -\sum_{r=1}^{\infty} \frac{x^r}{r}$ for $|x| $= -\log_{10}e \cdot \left(-\sum_{r=1}^{\infty} \frac{(1/n)^r}{r}\right) = \log_{10}e \cdot \sum_{r=1}^{\infty} \frac{n^{-r}}{r}$. Thus,the coefficient of $n^{-r}$ is $\frac{\log_{10}e}{r} = \frac{1}{r \log_e 10}$.

The coefficient of $n^{-r}$ in the expansion of $\log_{10}\left(\frac{n}{n-1}\right)$ is

Similar Questions

The value of $\log_e \left( 1 + ax^2 + a^2 + \frac{a}{x^2} \right)$ is

If $4\left[ {{x^2} + \frac{{{x^6}}}{3} + \frac{{{x^{10}}}}{5} + \dots} \right] = {y^2} + \frac{{{y^4}}}{2} + \frac{{{y^6}}}{3} + \dots$,then

$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}{1 \cdot 2} - \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} - \frac{1}{4 \cdot 5} + \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

Vedclass Test Series

Exam Paper Generator

Online Exam Module