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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.

Michael Hardy
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omar1810
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  • 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 03:05
  • 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 03:13
  • 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 03:16
  • 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 03:48

3 Answers3

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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
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Mhenni Benghorbal
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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 $x=-1.$

Bumblebee
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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}

Leucippus
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