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A book gave me the following series, and asked for which $r\in \mathbb{R}$ does it converge:

$$\frac{r}{2} + \frac{4r^2}{9} + \frac{9r^3}{28} + \frac{16r^4}{65} + \dotsb$$

I feel dumb because I can't even find the pattern in the denominator! It's not arithmetic (adding 7, then adding 19, then adding 37), looking at the prime factors doesn't seem to help ($2$, $3\cdot3$, $2\cdot2\cdot7$, $5\cdot 13$). Can someone see it?

Sorry, I'm sure it's obvious and I'm just missing it.

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I hate to be "overly technical to the point it is useless" but this book needs to give additional clues: for example that the numerator and denominator are polynomials or something along those lines. Since technically this could be any pattern that starts with those terms and therefore is a poor question (not your fault! The books fault!) – frogeyedpeas Jul 6 '13 at 1:46
@frogeyedpeas: No other clues are necessary: in problems of this kind there is a tacit assumption that the general term has a reasonable nice description. Students should of course learn at some point that a finite initial segment does nothing to determine an infinite sequence, but at this stage it’s much more important that they be able to spot simple numerical patterns. – Brian M. Scott Jul 6 '13 at 3:47
up vote 6 down vote accepted

The general term of the series is $$a_n=\frac {n^2r^n}{n^3+1}\sim_\infty \frac {r^n}{n}$$ so by the ratio test the Radius of convergence of the serie is $R=1$ and for $r=-1$ we can show that the series is convergent by the alternating test and it's a divergent series at $r=1$ and finally the interval of convergence is $[-1,1)$.

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I answered the question of the book. – user63181 Jul 6 '13 at 1:29
Well I respect your opinion even though the comment seems to contain a mockery but do I have to answer exactly the question and I do not have the right to add what seems to me useful for the reader? – user63181 Jul 6 '13 at 1:48

In the denominator, you want $$9=2^3+1$$ $$28=3^3+1$$ $$65=4^3+1$$

Thus, the general term is $$x_n=\frac{n^2}{n^3+1}r^n$$

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Thanks very much! – Eric Auld Jul 6 '13 at 1:04
Nice answer! Literally! – amWhy Jul 6 '13 at 3:24

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