I'm trying to calculate the sum of to infinity of the geometric series $\displaystyle 99,33,11,... $
$\displaystyle S_{\infty} = \frac{a}{1-r} $
$\displaystyle S_{\infty} = \frac{99}{1-3} = -\frac{99}{2}$
Is that right?
I'm trying to calculate the sum of to infinity of the geometric series $\displaystyle 99,33,11,... $
$\displaystyle S_{\infty} = \frac{a}{1-r} $
$\displaystyle S_{\infty} = \frac{99}{1-3} = -\frac{99}{2}$
Is that right?
Hello, Captcha!
Your $\displaystyle r$ is wrong . . .
Each term is divided by 3: .$\displaystyle r \,=\,\tfrac{1}{3}$Calculate the sum of the geometric series: .$\displaystyle 99,\;33,\;11,\;\hdots$
. . $\displaystyle S \;=\; \frac{a}{1-r} \;=\; \frac{99}{1-\frac{1}{3}} \;=\; \frac{99}{\frac{2}{3}} \;=\;\frac{297}{2}$
No, it isn't. For one thing, you should have noticed yourself that you are summing a series of positive numbers but the sum you got is negative. For another, along with that formula for the sum of an infinite geometric series, you should have learned that it is only valid for $\displaystyle -1\le r< 1$.
The first term of a geometric series is a and the second is ar. Here, a= 99, alright, but (99)(3) is NOT 33. Your r is wrong. 99r= 33 so what is r?
Oh, blast! Soroban got in 5 minutes before me. I need to learn to type faster!
Ah, I somehow calculated $\displaystyle r$ from $\displaystyle 33r = 99$. Thank you.
Yeah. I should have.
Yeah. I was told that. Why is that, though? Why does this formula not work for $\displaystyle -1\le r< 1$? How could we calculate the sum if $\displaystyle r > 1$ or r $\displaystyle \le -1$ ?For another, along with that formula for the sum of an infinite geometric series, you should have learned that it is only valid for $\displaystyle -1\le r< 1$
$\displaystyle r = \frac{33}{99} = \frac{1}{3}$, of course. Thanks.The first term of a geometric series is a and the second is ar. Here, a= 99, alright, but (99)(3) is NOT 33. Your r is wrong. 99r= 33 so what is r?
HI
The infinite GP formula actually came from this :
$\displaystyle S_n=\frac{a(1-r^n)}{1-r}$
$\displaystyle =\frac{a-ar^n}{1-r}$
$\displaystyle =\frac{a}{1-r}-\frac{ar^n}{1-r}$
Now , as n approaches infinity
$\displaystyle \lim_{n\rightarrow \infty}S_n=\frac{a}{1-r}-\frac{ar^n}{1-r}$
Consider this : $\displaystyle (0.1)^{10}$ and $\displaystyle 2^{10}$ .. which is approaching 0 ? It would be $\displaystyle 0.1^{10} $ right . That's why in order for $\displaystyle \frac{ar^n}{1-r}$ to be 0 , the $\displaystyle ar^n$ has to first be 0 which would only happen if $\displaystyle |r|<1$ . If this condition isn't applied , then it would approach infinity instead of 0 .