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Value of $\sum\limits_n x^n$

Ive been studying Geometric series and Arithmetic series all day and have struggled to attempt these problems. The Question is to sum up these problems.

1) \begin{align}&\sum_{n=0}^\infty 3^{-n}\end{align} 2) \begin{align}&\sum_{n=2}^\infty 3^{-n}\end{align} 3) \begin{align}&\sum_{n=0}^{n+1} 6^{n}\end{align}

Is it correct to say they are not geometric series because for 1) r=3 so r>1? So what formula do I use on these problems? I'm struggling to find the formulas to use.

Thanks. Apologies if I have asked this question the wrong way.

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marked as duplicate by Aryabhata, t.b., J. M., Zev Chonoles Apr 29 '12 at 8:52

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Since $3^{-n} = (\frac{1}{3})^n$, numbers 1 and 2 will converge. Number 3 is also finite, since you are summing only finitely-many terms. –  Austin Mohr Apr 6 '12 at 17:40
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2 Answers 2

You overlooked the minus sign in the exponent. Your $r$ in 1) is $\frac{1}{3}$. See if you can do it now, otherwise ping me with a comment and I'll post some more help.

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Oh dear. I cant believe it. I'm devastated at that mistake. Thank you so much, I lost 4 hours researching this problem. –  Shane Apr 6 '12 at 17:45
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@Shane pats on Shane's shoulder I know what it feels like. : ) –  Matt N. Apr 6 '12 at 17:49
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A geometric series is defined by the fact that successive terms are in the same ratio, so all the series you are summing are geometric. Whether the sum converges or not, the series can still be geometric. $\begin{align}&\sum_{n=0}^\infty 3^{n}\end{align}$ is still an attempt to sum a geometric series, but the sum is not convergent in this case.

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