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Find all the real values of $x$ for which $\sum_1^{\infty}(x^n/n)$ converges.

I began with the ratio test to get $nx/(n+1)$ but I'm not sure where to go next. I think I'm supposed to use Leibniz's Theorem at some point?

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You should have $x^n$ in the denominator ? –  Amr Nov 25 '12 at 22:33
    
Yes, sorry that was a typo, I'll correct it. –  Mathlete Nov 25 '12 at 22:34
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1 Answer

$$a_n:=\frac{x^n}{n}\Longrightarrow \left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{x^{n+1}}{n+1}\frac{n}{x^n}\right|=|x|\frac{n}{n+1}\xrightarrow [n\to\infty]{}|x|$$

Thus, the series converges for $\,-1<x<1\,$ , and it's easy to see that it also converges for $\,x=-1\,$ (Leibnitz series)

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That doesn't really make it any clearer; would the last term not tend to 0 as n tended to infinity? Also, how is it 'easy to see' that it converges for x=-1? How have you used Leibniz's theorem? –  Mathlete Nov 25 '12 at 22:36
    
Are you familiar with the alternating series test, Mathlete? –  Gerry Myerson Nov 25 '12 at 22:37
    
Mathlete, when $n=1000000000000$, is $n/(n+1)$ close to zero? or is it close to $1$? –  Gerry Myerson Nov 25 '12 at 22:38
    
OK I see what @DonAntonio meant now. –  Mathlete Nov 25 '12 at 22:39
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