Hey, I'm currently in a dispute with another student that perhaps somebody could help me settle? The question is, "Is it possible for two distinct geometric series to converge to the same value? If so, list such a pair. If not, state why not." The other student says that it is impossible--I'm still waiting to hear his reasoning as to why. I, however, am skeptical, and I think that it is possible, but I'm having trouble coming up with an example. Anybody out there able to help me out?

Question

anonymous · Answer

The series is infinite I supposed.

anonymous · Answer

Sum=a/(1-r)
so we can vary a and r to make the sum of the 2 series equal.
In short, yes can make the 2 geometric series converge to the same sum.

anonymous · Answer

I thought so! But which would be easier to vary? I'm trying to ensure that my example makes sense here. My first thought would be to vary a and keep r constant, but I'm not entirely sure that that would be feasible.

ganeshie8 · Answer

$$\large  \dfrac{a_1}{1-r_1} =  \dfrac{a_2}{1-r_2}$$

ganeshie8 · Answer

$$\large a_1(1-r_2) =  a_2(1-r_1)$$

ganeshie8 · Answer

$$\large a_1 - a_2 =   a_1r_2 -  a_2 r_1$$

ganeshie8 · Answer

say,  $a_1 = 2$, $a_2 = 1$ :

$$\large 1 =   2r_2 -  r_1$$

$$\large r_1 =   2r_2 -1$$

ganeshie8 · Answer

plugin $r_2 =   1/4$, you would get $r_1 =   -1/2$

anonymous · Answer

But must ensure that both |r|'s are less than 1

ganeshie8 · Answer

Also looks like $r_2 = 1/2$ produces nonsense

anonymous · Answer

Okay, I've found a suitable example pair using the method implied by the formula ganshie uses; if, say, I let the following hold true:

$$a_{1}=1, a_{2}=3, r_{1}=5$$

and I try solving for $$r_{2}$$, then I get r(2)=13. Checking back to make sure that my reasoning is true, I find that both sums come to -1/4=-1/4, thus making 

$$\sum_{0}^{\infty} 5^n = \sum_{0}^{\infty} 3(13)^n$$

anonymous · Answer

Thanks, everyone! I've got what I need. Thank you for your time.

ganeshie8 · Answer

hey the geometric series will not converge for $\large   |r| >  1$

ganeshie8 · Answer

So choose  some value for r between (-1, 1)

ganeshie8 · Answer

$$\large    \sum_{0}^{\infty} 5^n \ne  \sum_{0}^{\infty} 3(13)^n \color{red}{\ne}  -\frac{1}{4}$$

ganeshie8 · Answer

if we add up powers of 5, how can we ever arrive at -1/4 ?

anonymous · Answer

You're right. Revised accordingly. set up r(1)= 1/5 and a(1)=2 instead. Got more agreeable answer. Thanks for pointing that out before I could screw it up.

ganeshie8 · Answer

looks good :)