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May 29th, 2015, 03:29 AM   #1
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My poor brain can't handle this sequence

Hi! I am having trouble with this sequence. I found the general term An but I dont know how to find the sum. Can someone help me?
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May 29th, 2015, 04:04 AM   #2
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Let $$\begin{aligned}
s &= \phantom{2 + } \frac23 + \frac49 + \frac6{27} + \cdots \\
&= \phantom{2 + } {2 \over 3} + {4 \over 3^2} + {6 \over 3^3} + \cdots \\
3s &= 2 + {4 \over 3} + {6 \over 3^2} + {8 \over 3^3} + \cdots \\
3s - s &= 2 + {2 \over 3} + {2 \over 3^2} + {2 \over 3^3} + \cdots \\
&= 2\left( 1 + \left(\frac13\right) + \left(\frac13\right)^2 + \left(\frac13\right)^3 + \cdots\right)
\end{aligned} $$
You should be able easily to evaluate the last bracket to leave yourself with a simple piece of algebra to find $s$.
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May 29th, 2015, 05:09 AM   #3
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That's the infinite sum. The finite sum is
$$
S=r+r^2+\cdots+r^n
$$
which satisfies
$$
rS=r^2+r^3+\cdots+r^{n+1}
$$
so
$$
S-rS=r-r^{n+1}
$$
and thus
$$
S=\frac{r-r^{n+1}}{1-r}.
$$

I used "$r$" in place of "2/3" for generality.
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May 29th, 2015, 10:37 AM   #4
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Hello, matisolla!

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Last edited by skipjack; May 29th, 2015 at 03:32 PM.
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