# Thread: The sum (to infinity) of a geometric series.

1. ## The sum (to infinity) of a geometric series.

I'm trying to calculate the sum of to infinity of the geometric series $99,33,11,...$

$S_{\infty} = \frac{a}{1-r}$

$S_{\infty} = \frac{99}{1-3} = -\frac{99}{2}$

Is that right?

Your $r$ is wrong . . .

Calculate the sum of the geometric series: . $99,\;33,\;11,\;\hdots$
Each term is divided by 3: . $r \,=\,\tfrac{1}{3}$

. . $S \;=\; \frac{a}{1-r} \;=\; \frac{99}{1-\frac{1}{3}} \;=\; \frac{99}{\frac{2}{3}} \;=\;\frac{297}{2}$

I'm trying to calculate the sum of to infinity of the geometric series $99,33,11,...$

$S_{\infty} = \frac{a}{1-r}$

$S_{\infty} = \frac{99}{1-3} = -\frac{99}{2}$

Is that right?
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 $-1\le r< 1$.

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!

4. Originally Posted by Soroban

Your $r$ is wrong . . .

Each term is divided by 3: . $r \,=\,\tfrac{1}{3}$

. . $S \;=\; \frac{a}{1-r} \;=\; \frac{99}{1-\frac{1}{3}} \;=\; \frac{99}{\frac{2}{3}} \;=\;\frac{297}{2}$

Ah, I somehow calculated $r$ from $33r = 99$. Thank you.
Originally Posted by HallsofIvy
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.
Yeah. I should have.
For another, along with that formula for the sum of an infinite geometric series, you should have learned that it is only valid for $-1\le r< 1$
Yeah. I was told that. Why is that, though? Why does this formula not work for $-1\le r< 1$? How could we calculate the sum if $r > 1$ or r $\le -1$ ?
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?
$r = \frac{33}{99} = \frac{1}{3}$, of course. Thanks.

Yeah. I was told that. Why is that, though? Why does this formula not work for $-1\le r< 1$? How could we calculate the sum if $r > 1$ or r $\le -1$ ?
.
HI

The infinite GP formula actually came from this :

$S_n=\frac{a(1-r^n)}{1-r}$

$=\frac{a-ar^n}{1-r}$

$=\frac{a}{1-r}-\frac{ar^n}{1-r}$

Now , as n approaches infinity

$\lim_{n\rightarrow \infty}S_n=\frac{a}{1-r}-\frac{ar^n}{1-r}$

Consider this : $(0.1)^{10}$ and $2^{10}$ .. which is approaching 0 ? It would be $0.1^{10}$ right . That's why in order for $\frac{ar^n}{1-r}$ to be 0 , the $ar^n$ has to first be 0 which would only happen if $|r|<1$ . If this condition isn't applied , then it would approach infinity instead of 0 .