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 $\displaystyle 99,33,11,...$

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

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

Is that right?

Your $\displaystyle r$ is wrong . . .

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

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

3. Originally Posted by Captcha 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?
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!

4. Originally Posted by Soroban Your $\displaystyle r$ is wrong . . .

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

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

Ah, I somehow calculated $\displaystyle r$ from $\displaystyle 33r = 99$. 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 $\displaystyle -1\le r< 1$
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 $\displaystyle \le -1$ ?
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?
$\displaystyle r = \frac{33}{99} = \frac{1}{3}$, of course. Thanks.

5. Originally Posted by Captcha 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 $\displaystyle \le -1$ ?
.
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 .

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