There's something about a geometric series that makes it easily verifiable. Series like $\sum\frac{10^n}{9^n}$ or $\sum\frac{1}{2^n}$ aren't too bad; the variables $n$ are simple and easily reachable, and the fractions are not complex. But I'm having trouble with a series that looks somewhat different:


Its sequence converges, so I know I can apply the learned methods. The first thing I did was extract a constant from the sequence. So I go from the original sequence, which is:

$$a_n = \{\frac{2}{729}, \frac{4}{59049}, \frac{8}{478296}, \frac{16}{387420489}\}$$


$$a_n = \frac{2}{9}(\frac{1}{81}, \frac{2}{6561}, \frac{4}{531441}, \frac{8}{43046721})$$

I figured out the new sequence as: $\frac{2^n}{9^{2n}}$, and after the simplifying the constants, I was able to recreate the series in an almost geometric form of $ar^{n-1}$, with $\frac{1}{9}$ as $a$ and $\frac{2^n}{9^{2n}}$ as kind of my $r$. Right now, I have this:


This is sort of my dilemma. Having the $2n$ in the denominator is a serious issue; it prevents me from creating an $ar^{n-1}$ formula, and I need an $ar^{n-1}$ formula if I want to test the convergence of this series, at least with the methods I've learned so far. So I'm quite stuck.

Did something go wrong in my calculations? How can I turn this into the proper formula so I can test the series' convergence? Any help is appreciated.

Much Thanks,






  • $\begingroup$ --Knocks Head on wall-- Really? Wow. Thanks. :) $\endgroup$ – Zolani13 Apr 21 '12 at 22:42

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