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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 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 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 $s$. Thanks from Country Boy and matisolla
 May 29th, 2015, 05:09 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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. Thanks from matisolla
May 29th, 2015, 10:37 AM   #4
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Hello, matisolla!

Quote:
 $\text{Find the sum to the }n^{th}\text{ term: }\:\frac{2}{3}\,+\,\frac{4}{9}\,+\,\frac{6}{27} \,+\,\cdots\,+\,\frac{2n}{3^n}$

$\begin{array}{cccccc}
\text{W\!e have:} & S &=& \frac{2}{3}\,+\.\frac{4}{9}\,+\,\frac{6}{27}\,+\, \frac{8}{81}\, +\,\cdots\,+\,\frac{2n}{3^n}\;\;\;\;\;\;\;\;\; \;\;\;\;\\ \\
\text{Multiply by }\frac{1}{3}: & \;\frac{1}{3}S &=& \;\;\;\;\; \frac{2}{9}\,+\,\frac{4}{27}\,+\,\frac{6}{81}\,+\, \cdots\,+\,\frac{2(n-1)}{3^n} \,+\,\frac{2n}{3^{n+1}} \\ \\
\text{Subtract:} & \frac{2}{3}S &=& \frac{2}{3}\,+\,\frac{2}{9}\,+\,\frac{2}{27} \,+\, \frac{2}{81}\,+\, \cdots\,+\,\frac{2}{3^n} \,-\,\frac{2n}{3^{n+1}}\;\;\;\; \\ \\
\text{W\!e have:} & \frac{2}{3}S &=& \; \frac{2}{3} \underbrace{\left(1\,+\,\frac{1}{3}\,+\,(\frac{1}{ 3})^2\,+\,(\frac{1}{3})^3\,+\, \cdots \,+\,(\frac{1}{3})^{n-1}\right)}_{ \text{geometric series}}\,-\,\frac{2n}{3^{n+1}}
\end{array}$

$\text{The series has the sum: }\:\frac{1\,-\,(\frac{1}{3})^n}{1\,-\,\frac{1}{3}} \:=\:\frac{3^n\,-\,1}{3^n(\frac{2}{3})}$

$\text{Then we have: }\:\frac{2}{3}S \;=\;\frac{3^n\,-\,1}{3^n}\,-\,\frac{2n}{3^{n+1}} \;=\;\frac{3^{n+1}\,-\,3\,-\,2n}{3^{n+1}}$

$\text{Hence: }\:S \;=\;\frac{3}{2}\,\left(\frac{3^{n+1}\,-\,3\,-\,2n}{3^{n+1}}\right)$

$\text{Therefore: }\:S \;=\;\frac{3^{n+1}\,-\,3\,-\,2n}{2\,\cdot\,3^n}$

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

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