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May 29th, 2015, 03:29 AM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  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?

May 29th, 2015, 04:04 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 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 $s$. 
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 $$ SrS=rr^{n+1} $$ and thus $$ S=\frac{rr^{n+1}}{1r}. $$ I used "$r$" in place of "2/3" for generality. 
May 29th, 2015, 10:37 AM  #4  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, matisolla! Quote: Last edited by skipjack; May 29th, 2015 at 03:32 PM.  

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