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I have the following expression that I'm trying to simplify: $$\frac{1}{3} + \cdots+\frac{1}{3^{n-1}} +\frac{1}{3^n}.$$

This looks like a summation of $1$ to $n$ but in different terms. Can someone please explain how to simplify this/combine it into one term?

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  • $\begingroup$ Geometric series: $\sum_{i=1}^n 1/3^i = \frac{1-\frac{1}{3^{n+1}}}{1-\frac{1}{3}}-1$. $\endgroup$ – J.R. Feb 17 '14 at 19:47
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Let $p=\frac{1}{3}$ and let our sum be $S$. Note that $$pS=p^2+p^3+\cdots+p^{n+1}.$$ Calculate $S-pS$. There is a tremendous amount of cancellation, and we get $$S(1-p)=p-p^{n+1}.$$ Thus $S=\frac{p}{1-p}(1-p^n)$.

Remark: Exactly the same technique can be used to find a closed form for the sum of the finite geometric series $a+ar+\cdots +ar^{n-1}$, when $r\ne 1$. Ours was the case $a=r=\frac{1}{3}$.

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Hint. If $S = \frac 1 3 ^1 + \frac 1 3 ^2 + \dots$, then $S = \frac 1 3 (1+S)$. Also $\frac 1 3 ^1 + \frac 1 3 ^2 + \dots + \frac 1 3 ^n = S - \frac 1 3 ^nS$.

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