2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @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
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.