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October 26th, 2017, 03:36 AM   #1
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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.

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October 26th, 2017, 04:24 AM   #2
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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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