Question

Question about infinite series.

Answer 1

recall: \[a^{-k}=\left(\frac{1}{a}\right)^k\]

Answer 2

This is a special series.

Answer 3

Special is good.

Answer 4

define the innards as the ratio and work the geometric sum to equate to 8 is my thought

Answer 5

It is geometric.

Answer 6

a= 1+c and r= 1/1+c?

Answer 7

good

Answer 8

I believe when \(0\leq r<1\) we have: \[
\sum_{n=0}^\infty r^n = \frac{1}{1-r}
\]

Answer 9

And so: \[
\sum_{n=1}^\infty r^n = \frac{1}{1-r} - r^0 = \frac{1}{1-r}-1
\]

Answer 10

r=0?

sum of 0^n = 1 not buying it :)

Answer 11

@wio where did you get the r^0?

Answer 12

\(\color{blue}{\text{Originally Posted by}}\) @amistre64
r=0?

sum of 0^n = 1 not buying it :)
\(\color{blue}{\text{End of Quote}}\)
Why not? \(0^0\) is an indeterminate form.

Answer 13

It can be whatever we want it to be

Answer 14

not gonna believe it ... cant make me .... nah nah nah nah ....

Answer 15

\(\color{blue}{\text{Originally Posted by}}\) @misspacgirl14
@wio where did you get the r^0?
\(\color{blue}{\text{End of Quote}}\)
It is the first term in the sequence. Since we started at \(n=0\), I had to subtract \(r^0\) from both sides for \(n=1\).

Answer 16

\[S=r+r^2+r^3+r^4+....\]
\[S/r=1+r+r^2+r^3+....\]

Answer 17

S/r = S+1 as well

Answer 18

Okay but I'm still confused about how to go about this problem..

Answer 19

we have r + r^2 + r^3 + r^4 + ....

right?

Answer 20

(1+c) / 1- (1/(1+c)) =8 I have this so far...

Answer 21

ok

Answer 22

S/r = 1 + r + r^2 + r^3 + r^4

S/r = 1/(1-r)

S = r/(1-r)

and r = 1/(1+c)

Answer 23

your top parts is bad, the rest is just simple algebra

Answer 24

\(a_1 = \dfrac{1}{1+c}\)

\(r = \dfrac{1}{1+c}\)

Answer 25

Look, we are basically saying that: \[
\sum_{n=1}^\infty r^n = \frac{1}{1-r}-1
\]let \(r=1/(1+c)\)

Answer 26

r/(1-r) = 8

r= 8(1-r)

0 = 8(1-r) - r

0 = 8-9r

when r=8/9

Answer 27

Or another way would be to solve (1/ (1+c)) / (1 - (1/(1+c))) = 8?

Answer 28

yes that looks more straightforward

Answer 29

you could but its a bother, keep things simple, solve for r in terms of the original setup then sub it in

Answer 30

Okay that makes more sense to me :) Thank you all your help everyone. I wish I could give you guys all medals.