Answer to Question #161771 in Calculus for Vishal

Question

Show that ∞n=1[(−1)^(n−1)].[(x^n)/(n)] converges if |x| < 1

Answer 1

$\sum _{n=1}^{\infty \:}\left(-1^{n-1}\right)x^n\frac{1}{n}$

$\mathrm{If\:there\:exists\:an\:}N\mathrm{\:so\:that\:for\:all\:}n\ge N,\:\quad a_n\ne 0\mathrm{\:and\:}\lim _{n\to \infty }|\frac{a_{n+1}}{a_n}|=L:$

$\mathrm{If\:}L<1\mathrm{,\:then\:}\sum a_n\mathrm{\:converges}$

$\mathrm{If\:}L>1\mathrm{,\:then\:}\sum a_n\mathrm{\:diverges}$

$\mathrm{If\:}L=1\mathrm{,\:then\:the\:test\:is\:inconclusive}$

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

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

$\frac{\left(-1^{\left(n+1\right)-1}\right)x^{n+1}\frac{1}{n+1}}{\left(-1^{n-1}\right)x^n\frac{1}{n}}$

$\frac{-1^{n+1-1}x^{n+1}\frac{1}{n+1}}{-1^{n-1}x^n\frac{1}{n}}$

$\frac{-\frac{1}{n+1}x^{n+1}}{-1^{n-1}\cdot \frac{1}{n}x^n}$

$\frac{x^{n+1}\frac{1}{n+1}}{1^{n-1}x^n\frac{1}{n}}$

$\frac{\frac{x^{n+1}}{n+1}}{1^{n-1}\cdot \frac{1}{n}x^n}$

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

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

$\frac{nx}{n+1}$

$\lim _{n\to \infty \:}\left(\left|\frac{nx}{n+1}\right|\right)$

$=\left|x\right|\cdot \lim _{n\to \infty \:}\left(\left|\frac{n}{n+1}\right|\right)$

$\left|x\right|\cdot \:1$

$-1<x<1$

$\mathrm{Check\:convergence\:for\:}L=1$

$\lim _{n\to \infty \:}\left(\frac{1}{n}\right)=0$

$\mathrm{Therefore,\:the\:convergence\:interval\:of\:}\sum _{n=1}^{\infty \:}\left(-1^{n-1}\right)x^n\frac{1}{n}\mathrm{\:is\:} -1\le \:x<1$