absolute convergence of an infinite series

Show that the series:

$$1 + \frac{x}{1\cdot 2} + \frac{x^2}{2\cdot 3} + \frac{x^3}{3\cdot 4} + \cdots$$

is absolutely convergent when $-1< x <+1$.

I've been trying to prove this however am having difficulty when $x = 1$ where it would seem to converge as a telescoping series. Please any help would be appreciated thanks.

• If it says "when $-1<x<+1$, that excludes the case $x=1$. However, if you can show it converges as a telescoping sequence, then, since every term is non-negative, that shows it converges absolutely, so you're done. – Michael Hardy Sep 12 '14 at 3:05
• You are asked to show the series is "convergent when $-1<x<1$", not "convergent only when $-1<x<1$". So it doesn't mean you have to show it is divergent when $x=1$, it just means that convergence or divergence when $x=1$ is not part of the question. – David Sep 12 '14 at 3:13
• Most of the questions of this type in the book have included in the interval x = 1 and x = -1 also. So with the question excluding those cases I'm assuming that I must have gone wrong somewhere! How would you go about proving this question? – omar1810 Sep 12 '14 at 3:16
• Just for your curiosity, the series coincides with the Taylor expansion of $$2+\left(\frac{1}{x}-1\right) \log (1-x)$$ – Claude Leibovici Sep 12 '14 at 3:48

Just write the series as

$$S = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n(n+1)}.$$

Now if $x=1$ you can use telescoping techniques

$$\sum_{k=1}^{\infty} \frac{1}{k(k+1)}.$$

You need to consider the partial sum

$$S_n = \sum_{k=1}^{n} \left( \frac{1}{k}- \frac{1}{k+1} \right)$$

and then take the limit as $n$ goes to infinity.

• @Leucippus: Thanks for the edit. I really appreciate it. – Mhenni Benghorbal Sep 12 '14 at 3:51

For $-1<x<1$ you can show that using ratio test.
When $n=1$ the $n^{th}$ term of the series become $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ Then you can apply the telescoping techniques from here. Note that the series $$\frac{(-1)^n}{n(n+1)}$$ is absolutely convergent. Therefore your series is also convergent at $x=-1.$

The value of the series is \begin{align} 1 + \sum_{k=1}^{\infty} \frac{ x^{k} }{ k(k+1) } = \frac{1-x}{x} \, \ln(1-x). \end{align}