Working out which x for the Conditional Convergence of a Series

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?

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 $$x=-2$$.

• Hello I see what you've done here but I'm slightly confused, should it not be $$R=\lim\limits_{n\to\infty}\frac{|a_{|n+1}|}{|a_n|} = ?$$ – king Mar 1 at 2:47
• @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 $x$. – Haris Gušić Mar 1 at 2:56
• @king Did you understand this? Was it of any help? – Haris Gušić Mar 1 at 20:11
• Yes it makes complete sense thank you – king Mar 3 at 16:14
• @king If this answer answered your question, you can mark it as accepted. – Haris Gušić Mar 3 at 22:51

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 $$-\frac{1}{2}\ln \frac{1}{2}.$$