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How do I express $\frac{1}{3^n}$, or $3^{-n}$ as a typical geometric series of the form: $ar^{n-1}$ ?

This is a small part of a larger question, in which I'm trying to find the convergence of $\sum_{n=0}^\infty \frac{1+2^n}{3^n}$.

I broke the series into two separate series', $\sum_{n=0}^\infty \frac{1}{3^n} + \sum_{n=0}^\infty (\frac23)^n $.

However I am stuck trying to represent the first term as a geometric series. I know that it is a geometric series only because Wolfram Alpha told me so here.

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    $\begingroup$ Not exactly sure what is confusing you here. $\frac{1}{3^n} = (\frac{1}{3})^n = 0.333...^n$ which obviously converges since $0.333... < |1|$. $\endgroup$ – SilverSlash Jan 8 '16 at 15:13
  • $\begingroup$ Thanks for the tip here. I oversaw the $\frac{1}{3^n} = (\frac{1}{3})^n$. I forgot (stupidly) that $1^n$ is, of course, just $1$. $\endgroup$ – k4kuz0 Jan 8 '16 at 15:18
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HINT: Notice, $$\frac 1{3^{n}}=\frac{1}{3\cdot 3^{n-1}}=\frac{1}{3}\cdot \frac{1}{3^{n-1}}=\frac{1}{3}\cdot \left(\frac{1}{3}\right)^{n-1}$$ comparing with typical form of G.P. $ar^{n-1}$, one should have $a=\frac 13$ & $r=\frac 13$

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  • $\begingroup$ Thank you. I feel stupid for not seeing it. $\endgroup$ – k4kuz0 Jan 8 '16 at 15:16
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In the expression $a r^{n-1}$, we think of $a$ as representing the first term of the sequence, and $r$ as the ratio between common terms.

Examining the sequence given, we note that:

  • the first term is $\frac{1}{3^1}$
  • the ratio of terms is $\frac{\frac{1}{3^{n+1}}}{\frac{1}{3^n}}=1/3$

So, $a=r=\frac{1}{3}$.

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