October 26th, 2017, 02:36 AM  #1 
Member Joined: Aug 2013 Posts: 41 Thanks: 3  Limiting sum
Suppose a Geometric series has a limiting sum = 2. Find the values of the first term 'a'. Its rather easy to show that 0<a<4, but can a = 2? If a = 2, then r =0. Is this still a geometric series? the series would look like this: 2+0+0+0....., so essentially a useless series, but is it geometric? Now, with a GP, T2/T1 = T3/T2. In my series above, T2/T1 = 0/2 = 0. But T3/T2 = 0/0 = undefined. But, a formal definition, gives T3/T2 = ar^2/ar = ar/a, when r =0, then ar/a = 0 which isn't undefined. Just curious if anyone has any thoughts on this. Hope the Math Forum community is doing well, just realized I haven't logged in for about 4 years. Eva 
October 26th, 2017, 03:24 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
A geometric series is of the form $\displaystyle \sum_{n=0}^\infty ar^n$ for given number a and r and its "limiting sum" (I would say just "sum") is given by $\displaystyle \frac{a}{1 r}$. If that "limiting sum" is 2 then we have $\displaystyle \frac{a}{1 r}= 2$ so $\displaystyle a= 2(1 r)$. In order that the "limiting sum" exist, we must have $\displaystyle 1< r< 1$ so that $\displaystyle 1< r< 1$, $\displaystyle 0< 1 r< 2$, and $\displaystyle 0< 2(1 r)< 4$. a, the first term in the series, can be any number between 0 and 4. Yes, r can be 0 so a can be 1. In that case, as you say, the "geometric sum" is just 2+ 0+ 0+ .... 

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