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I'v got roughly half way through this question:

For (fixed) x which is an element of the real numbers, consider the series

$\sum_{n=1}^\infty \frac{x^{n-1}}{2^nn} $

For which x does this series converge? For which x is the series conditionally convergent?

So far I've managed to deduce that for x=-1, the series is in oscillating harmonic form i.e.

$\sum_{n=1}^\infty \frac{x^{n-1}}{2^nn} = \sum_{n=1}^\infty \frac{1}{2^nn} * (-1)^{n-1} $

here $a_n$ = $\frac{1}{2^nn}$ where as n approaches infinity, $\frac{1}{2^nn}$ approaches 0. Therefore the series converges according to the alternating series test.

For all other values between -1 and 1, the series must be convergent as $a_n$ is decreasing.

Therefore I concluded that the series is absolutely convergent in the interval [-1,1].

Thats what I got so far but I'm quite uncomfortable with this area. Would anyone mind correcting/helping?

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You are right that the series converges absolutely on that interval. But that interval can be extended even further. And you cannot say that the series converges for $x \in (-1,1)$ just because $a_n$ is decreasing.

The radius of convergence can be obtained by the D'Alembert convergence test. We get that

$$R=\lim\limits_{n\to\infty}\frac{|a_n|}{|a_{n+1}|} = 2$$

So the series converges absolutely on $(-2,2)$. Let's look at the boundary points.

  1. For $x=2$ we have that the general term is $$\frac{2^{n-1}}{2^n n} = \frac{1}{2n}$$

    The series becomes a harmonic series which is divergent.

  2. For $x=-2$ we have that the general term is $$\frac{(-1)^{n-1}2^{n-1}}{2^n n} = \frac{(-1)^{n-1}}{2n}$$ and the series converges conditionally by the alternating series test, but not absolutely.

So, we have convergence on $[-2,2)$ with absolute convergence on $(-2,2)$ and conditional convergence for $x=-2$.

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  • $\begingroup$ Hello I see what you've done here but I'm slightly confused, should it not be $$R=\lim\limits_{n\to\infty}\frac{|a_{|n+1}|}{|a_n|} = ?$$ $\endgroup$ – king Mar 1 '19 at 2:47
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    $\begingroup$ @king No, my formula is for the radius of convergence. The $R$ I was refering to is the greatest $R$ s.t. the series converges on $(-R,R)$. By d'Alembert's test, for the series to converge we need that $$\lim\limits_{n\to\infty}\frac{|a_{n+1}||x|}{|a_n|} = |x| \lim\limits_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} < 1$$ From here you get that $$|x| < \left(\lim\limits_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}\right)^{-1} = \lim\limits_{n\to\infty} \frac{|a_n|}{|a_{n+1}|} = R$$ which is equivalent with $x \in (-R, R)$. Look up the radius of convergence. The $a_n$ I was using here doesn't contain $x$. $\endgroup$ – Haris Gušić Mar 1 '19 at 2:56
  • $\begingroup$ @king Did you understand this? Was it of any help? $\endgroup$ – Haris Gušić Mar 1 '19 at 20:11
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    $\begingroup$ Yes it makes complete sense thank you $\endgroup$ – king Mar 3 '19 at 16:14
  • $\begingroup$ @king If this answer answered your question, you can mark it as accepted. $\endgroup$ – Haris Gušić Mar 3 '19 at 22:51
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Let's calculate the value of the sum:

$S(x)=\sum\limits_{n=1}^\infty \dfrac{x^{n-1}}{2^nn}\tag1$

Using that $\frac{x^n}{n}=\int\limits_0^x x^{n-1}dx$ and replace the order of the summation and integration:

$\frac{1}{x}\int\limits_0^x \frac{1}{x}\sum\limits_{n=1}^\infty\big( \frac{x}{2}\big)^{n}dx\tag2$

$\sum\limits_{n=1}^\infty\big( \frac{x}{2}\big)^{n}$ is geometric series, convergent if $|\frac{x}{2}|\lt1$

Perform the summation then integration we get:

$S(x)=\frac{1}{x}\ln\frac{2}{2-x}\tag3$

So the series is absolute convergent if $|{x}|\lt2$

Regardnig the original series in $S(-2)$ convergent and has the value $-\frac{1}{2}\ln \frac{1}{2}.$

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