# $\sum_{n=1}^{\infty}\frac{x^n}{n}$ converges for $|x|<1$

I wish to show $\sum_{n=1}^{\infty}\frac{x^n}{n}$ converges for $|x|<1$, preferably using the comparison test.

I have done it using the limit ratio test, then having to show that $\lim_{n\to\infty}\frac{xn}{n+1}$ converges. To do this I tried factoring out $n$ to get:

$\lim_{n\to\infty}\frac{x}{1+1/n}$

which converges to x (if that technique is even a sure method), which is less than 1.

However I do not wish to use the limit ratio test, but rather I would like to use the comparison test, which I have not had much luck with. Thanks in advance for any hints.

• remove the $n$ from the denominator and compare with geometric series – Mirko Mar 23 '15 at 1:18
• Oh wow, I'm an idiot. Thank you so much! – anakhro Mar 23 '15 at 1:18
• Just so you know, what you did with the limit ratio test is valid. – graydad Mar 23 '15 at 1:19
• A nice follow up: show that $\sum n^\alpha x^n$ converges for any real $\alpha$ when $|x|<1$. – Robert Wolfe Mar 23 '15 at 1:22

$a_n = \dfrac{x^n}{n} \Rightarrow |a_n| = \dfrac{|x|^n}{n} < |x|^n$, and the geometric series $\displaystyle \sum_{n=1}^\infty |x|^n$ converges since $|x| < 1$, thus by comparison test the original series converges absolutely hence converges.