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I like to determine where the following power series converges.

$$\sum_{k=1}^\infty \dfrac{x^k}{k} $$

Since the harmonic series diverges, I think that the series would converge if I make the numerator small by forcing $|x|<1$, but I cannot rigoroulsy show where the series converges. How should I approach this problem?

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  • $\begingroup$ The harmonic series diverges or converges? $\endgroup$ – Conrado Costa Sep 8 '15 at 21:04
  • $\begingroup$ The harmonic series diverges $\endgroup$ – LASV Sep 8 '15 at 21:04
  • $\begingroup$ @ConradoCosta Sorry for the confusion. I meant diverges $\endgroup$ – mononono Sep 8 '15 at 21:05
  • $\begingroup$ @LASV Sorry for the confusion. I meant diverges $\endgroup$ – mononono Sep 8 '15 at 21:05
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    $\begingroup$ No problem. It might help to notice that the series above is the antiderivative of $\frac{1}{1-x}$ inside the unit disc and the antiderivative has the same radius of convergence as the original series. $\endgroup$ – LASV Sep 8 '15 at 21:10
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$|x|\leq 1$ is a necessary condition for convergence. The harmonic series is divergent, while: $$ \sum_{k\geq 1}\frac{(-1)^k}{k}=-\log(2). $$ On the other hand, if $|x|<1$ the series is absolutely convergent.

It follows that the series is convergent for $-1\leq x<1$. For such values of $x$, it is not difficult to check that:

$$ \sum_{k\geq 1}\frac{x^k}{k}=-\log(1-x).$$

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Hint. You may use the ratio test, evaluating $$ \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}(x)}{a_n(x)}\right| $$ with $$ a_n(x)=\frac{x^n}n.$$

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To see that the series converges for $|x|<1$ note that

$$\sum_{k= 1}^\infty \frac{|x|^k}{k} \leq \sum_{k= 1}^\infty |x|^k < \infty$$

this means that $\forall \epsilon >0$ $\exists\, N \in \Bbb{N}$ such that

$$m,n> N \Rightarrow \bigg|\sum_{k = n}^m \frac{x^k}{k} \bigg|\leq \sum_{k= 1}^\infty |x|^k < \epsilon$$

So the series converges

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