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.


Answer for your last question:

1, Absolut convergence:


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:


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:


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


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