The sum to infinity of the given series frac{ | vedclass

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

Question

(A) The given series is $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ This can be rewritten as $\frac{1}{n} - \frac{(1/n)^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given series is $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ This can be rewritten as $\frac{1}{n} - \frac{(1/n)^2}{2} + \frac{(1/n)^3}{3} - \frac{(1/n)^4}{4} + \dots$ We know the logarithmic series expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ for $|x| Substituting $x = \frac{1}{n}$,we get $\log_e(1 + \frac{1}{n}) = \frac{1}{n} - \frac{(1/n)^2}{2} + \frac{(1/n)^3}{3} - \dots$ Therefore,the sum is $\log_e\left(\frac{n+1}{n}\right)$.

The sum to infinity of the given series $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ is

Similar Questions

If $\tanh ^{-1} x = a \log \left(\frac{1+x}{1-x}\right)$, $|x| < 1$, then $a$ 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 = $

$\frac{2}{1} \cdot \frac{1}{3} + \frac{3}{2} \cdot \frac{1}{9} + \frac{4}{3} \cdot \frac{1}{27} + \frac{5}{4} \cdot \frac{1}{81} + \dots \infty = $

If $0 < x < 1$,then $\frac{3}{2} x^{2} + \frac{5}{3} x^{3} + \frac{7}{4} x^{4} + \ldots$ is equal to:

The value of the series $x \log _e a + \frac{x^3}{3!} (\log _e a)^3 + \frac{x^5}{5!} (\log _e a)^5 + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module