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$$\sum\limits_{n=1}^∞ (n^2 + \frac{n}{n+3}) = 1\frac{1}{4} + 4\frac{2}{5}+9\frac{3}{6} + 16\frac{4}{7} +25\frac{5}{8}...$$

I can't seem to figure out what the common ratio for this series is, not that it's non-existent, probably just that i'm not skilled enough, or just blind, i've tried what worked with other simple series, $\frac{a^n+1}{a^n}$, but in this case it didn't work, thank you in advance.

colormegone
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Sam Chahine
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  • A geometric series has the form $ \ \Sigma_{n=n_{init}}^{n_{term}} \ r^{n} \ , $ so that the base $ \ r \ $ in the exponential term is a constant and the exponent carries the index. – colormegone Jan 14 '14 at 04:44
  • I suppose rhat the "+" sign in the expression is "*". Otherwise, the rhs does not match the lhs. – Claude Leibovici Jan 14 '14 at 07:09

1 Answers1

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The series is not geometric, so has no common ratio.

For an easy way to see this, note that (labeling the $n$th term by $a_n$) we have $$\frac{a_2}{a_1}=\cfrac{\frac{22}{5}}{\frac54}=\frac{88}{25},$$ while $$\frac{a_3}{a_2}=\cfrac{\frac{19}2}{\frac{22}5}=\frac{95}{44}.$$ These particular ratios are not equal, and so....

Cameron Buie
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  • I see, does that mean there is no answer? Does it just tend to infinity? – Sam Chahine Jan 14 '14 at 00:40
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    The terms themselves grow without bound as $n$ does, so the series cannot possibly converge. In a convergent series, the terms tend toward $0$ as the index grows without bound. – Cameron Buie Jan 14 '14 at 00:43