Question

$\sum _{n=1}^{\infty }nx^n$

solve it

Answer 1

$\mathrm{Convergence\:Interval\:of\:}\sum _{n=1}^{\infty \:}nx^n:\quad -1<x<1$

$Steps$

$\sum _{n=1}^{\infty \:}nx^n$

$\mathrm{Use\:the\:Ratio\:Test\:to\:compute\:the\:convergence\:interval}:\quad -1<x<1$

$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{\left(n+1\right)x^{\left(n+1\right)}}{nx^n}\right|$

$\mathrm{Compute\:}\lim _{n\to \infty \:}\left(\left|\frac{\left(n+1\right)x^{\left(n+1\right)}}{nx^n}\right|\right):\quad \left|x\right|$

$\mathrm{The\:sum\:converges\:for\:}L<1\mathrm{,\:therefore\:solve\:}\left|x\right|<1$

$\mathrm{For}\:x=-1,\:\sum _{n=1}^{\infty \:}n\left(-1\right)^n:\quad \mathrm{diverges}$

$\mathrm{For}\:x=1,\:\sum _{n=1}^{\infty \:}n1^n:\quad \mathrm{diverges}$

$\mathrm{Therefore,\:the\:convergence\:interval\:of\:}\sum _{n=1}^{\infty \:}nx^n\mathrm{\:is\:}$

$-1<x<1$

Answer 2

Explanation:

$\sf \: \sum _{n=1}^{\infty }nx^n \\ \\ \: \sf \red{\sum _{n=1}^{\infty } \: n \: = x \times n}$