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 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
 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+ .... Thanks from evaeva

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