# a convergent geometric series...

• Mar 15th 2010, 02:16 PM
andrewt328
a convergent geometric series...
i need help finding a convergent geometric series that converges to 3.

i know the formula for a geometric sreries is the sum cr^n, and that if the absolute value of r < 1 it converges to zero...but i dont know how to make one converge to 3.

would this converge to 3? sum 3(2/3)^n, since 2/3 will go to zero and then the c value of 3 will make it converge to 3?

thanks for any help
• Mar 15th 2010, 02:22 PM
Plato
Quote:

Originally Posted by andrewt328
i need help finding a convergent geometric series that converges to 3.

Does $\displaystyle \sum\limits_{k = 0}^\infty {2 \cdot 3^{ - k} }$ work?
• Mar 15th 2010, 02:25 PM
jiboom
Quote:

Originally Posted by andrewt328
i need help finding a convergent geometric series that converges to 3.

i know the formula for a geometric sreries is the sum cr^n, and that if the absolute value of r < 1 it converges to zero...but i dont know how to make one converge to 3.

would this converge to 3? sum 3(2/3)^n, since 2/3 will go to zero and then the c value of 3 will make it converge to 3?

thanks for any help

3[(2/3)+(2/3)^2+.......]=2+2(2/3)+2(2/3)^2+2(2/3)^3+....

sum to infinity is a/1-r so we have 2/(1-2/3)=6

can you see how to amewnd your gp to get a sum of 3?
• Mar 15th 2010, 04:31 PM
andrewt328
Quote:

Originally Posted by Plato
Does $\displaystyle \sum\limits_{k = 0}^\infty {2 \cdot 3^{ - k} }$ work?

i dont think that will work...but thats not a geometrical series since it is to the -k...right?

and jiboom, since its c/1-r, wouldnt c be 3, and it be 3/1-(3/2)? and what does gp mean? im probably confused but i dont know
• Mar 16th 2010, 06:06 AM
General
Quote:

Originally Posted by andrewt328
i dont think that will work...but thats not a geometrical series since it is to the -k...right?

and jiboom, since its c/1-r, wouldnt c be 3, and it be 3/1-(3/2)? and what does gp mean? im probably confused but i dont know

But $\displaystyle 3^{-k}=\left( \frac{1}{3} \right)^k$. :o

For your problem, you want to find a two constant $\displaystyle a,r$ such that : $\displaystyle \frac{a}{1-r}=3$
choose any value for r such that $\displaystyle |r|<1$ (to make the series coverges) then put this value in the last equation and solve it for a.
Once you have $\displaystyle a,r$ you can put them in the general form of the geometric series to get the desired series.
• Mar 16th 2010, 10:53 AM
jiboom
Quote:

Originally Posted by andrewt328

and jiboom, since its c/1-r, wouldnt c be 3, and it be 3/1-(3/2)? and what does gp mean? im probably confused but i dont know

i wrote out the series, it is
2+2(2/3)+2(2/3)^2+2(2/3)^3+....

from there we can read off the first term (my a,presumably your c) as 2 and common ratio 2/3

gp mean geometric progression which is what your question is about.
• Mar 27th 2010, 06:08 PM
andrewt328
thanks for the help guys...