Summation 1 to infinity: $\sum\limits_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n(n+1)}$

$$\sum\limits_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n(n+1)}$$

Determine the values of x for which the given series

1. Converges absolutely
2. Converges conditionally
3. Diverges

On applying Ratio Test of Absolute Convergence we get $$|x|$$.

Which is convergent when $$|x| < 1$$ and divergent when $$|x| > 1$$.

How to analyze at $$|x| = 1$$?

For $$|x|=1$$, you have that $$|a_n| = \frac{1}{n(n+1)} \sim \frac{1}{n^2}$$ so the series is absolutely convergent.

1, Absolut convergence:

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

Using that $$\sum\limits_{n=2}^{\infty}\dfrac{x^{n-1}}{n}=\dfrac{1}{x}\sum\limits_{n=1}^{\infty}\dfrac{x^n}{n}-1$$ we get:

$$|S|=|1+(1-\frac{1}{x})\sum\limits_{n=1}^{\infty}\dfrac{x^n}{n}|=|(1-\frac{1}{x})Li_1(x)+1|=|\frac{1-x}{x}\ln(1-x)+1|=|\dfrac{\ln(1-x)}{\frac{x}{1-x}}+1|$$

In order to analize |S| at x=1 the L'Hospital rule is applied:

$$|S|_{x\rightarrow 1}\rightarrow \Big|\dfrac{\frac{1}{(1-x)}}{(\frac{1}{1-x})^2}+1\Big|\rightarrow 1$$

2, For the original sum can be use the same steps as before, the result is:

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

$$|S|_{x\rightarrow -1}\rightarrow -1$$