Finding the smallest possible value of the sum in infinite geometric sequence and another geometric sequence question

algebra-precalculusgeometric series

  1. The sum of an infinite geometric series is a positive number $S$, and the second term in the series is $1$. What is the smallest possible value of $S$?

For this I am thinking the $t_1, 1, t_2,…$

$r = \frac{1}{t_1} = \frac{t_2}{1}$ so that means $t_1t_2 =1$

$S = \frac{t_1}{1-t_2}$

I don't know how to get the smallest possible value for $S$ from here

  1. A geometric series with infinitely many terms has a sum of 2018. Now take each term in the original series and square it; this new series has a sum of 20180. The common ratio of the original series is $m/n$ where gcd$(m,n) =1$. Find $m+n$.

If the sum is 2018 then $a_1 = 2018(1-r)$. Then $a_1^2 = (2018-2018r)^2$

this is the value of the first term of the new series. so then the sum for that series would be $20180 = \frac{(2018-2018r)^2}{1-r}$

This is all I could come up with so far. Any help is appreciated, thank you.

Best Answer

1.$S=\frac{t_1}{1-t_2}=\frac{1/r}{1-r}=\frac{1}{r-r^2}$. Finding min for $S$ is the same as max for $\frac{1}{S}=r-r^2$, where the max is at $r=\frac{1}{2}$, leading to $S=4$ as the minimum.

  1. $20180=(2018)^2\frac{(1-r)^2}{1-r}$ So $10=2018(1-r)$ or $r=\frac{2008}{2018}=\frac{1004}{1009}$ where the numerator and denominator are relatively prime. So $m+n=2013$.
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