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

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 $|x| = 1$?

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

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Answer for your last question:

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:

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

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

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