I'm trying to find a general term for this series:

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

Without the one it's straightforward: $$\frac{x^n}{n(n+1)}$$

However I can't find a general term that includes the one. Please can you help.

  • $\begingroup$ @amWhy. This was a good one ! Cheers :-) $\endgroup$ – Claude Leibovici Sep 11 '14 at 11:48
  • $\begingroup$ So the question is only to find the general term rather than the sum? $\endgroup$ – hypergeometric Sep 11 '14 at 16:36
  • $\begingroup$ It was to prove that the series is absolutely convergent only for -1<x<1 however when x = 1 it seems to be convergent!? $\endgroup$ – omar1810 Sep 12 '14 at 2:44
  • $\begingroup$ @omar1810 Yes, since it is like $\frac1{n^2}$ for $x=1$. $\endgroup$ – AlexR Sep 12 '14 at 6:39

For a convenient way to write inline, I suggest $a_0(x) = 1, a_n(x) = \frac{x^n}{n(n+1)}, n\in\mathbb N$. If you have more space, you can also use $$a_n(x) = \begin{cases}\frac{x^n}{n(n+1)} & n >0\\ 1 & n=0\end{cases}$$ However, you won't get around the piecewise definition, since the general term contains a division by zero for $n=0$ and cannot be easily patched. If you accept $0^0 = 0$, you may put $(n+1-n^0)(n+1)$ in the denominator to obtain $$a_n(x) = \frac{x^n}{(n+1-n^0)(n+1)}$$ But you'd have to note $0^0 = 0$ to avoid confusion. Also, it's less readable.


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