# Thread: limits and geometric series

1. ## limits and geometric series

A) suppose that |r| < 1 find the limit:
Lim n->(infinity) (1+r+r^2+....+r^n)
B) If we let the infinite repeating decimal .9999 ...... stand for the limit
Lim n->(infinity) ( 9/10 + 9/(10^2) + ..... + 9/(10^n) ), show that .9999..... = 1

2. Originally Posted by luckyc1423
B) If we let the infinite repeating decimal .9999 ...... stand for the limit
Lim n->(infinity) ( 9/10 + 9/(10^2) + ..... + 9/(10^n) ), show that .9999..... = 1
let 0.999999.... = lim(n--> oo) (9/10 + 9/(10^2) + ..... + 9/(10^n) )
.....................= lim(n--> oo) 9/10 (1 + 1/10 + 1/(10^2) + ....)
.....................= lim(n--> oo) 9/10 * lim(n--> oo) (1 + 1/10 + 1/(10^2) + ....)
.....................= 9/10 *lim(n--> oo) (1 + 1/10 + 1/(10^2) + ....)

Note that (1 + 1/10 + 1/(10^2) + ....) is a geometric series with a = 1 and r = 1/10 (which means |r| < 1). so it's infinite sum is given by a/(1 - r)

......................= 9/10 * 1/(1 - 1/10) = 9/10 * 1/(9/10) = 9/10 * 10/9 = 1

3. Originally Posted by luckyc1423
A) suppose that |r| < 1 find the limit:
Lim n->(infinity) (1+r+r^2+....+r^n)
note that 1+r+r^2+....+r^n is a geometric series with a = 1 and r = r, so it's infinite sume is given by a/(1 - r) when |r|< 1

so Lim n->(infinity) (1+r+r^2+....+r^n) = 1/(1 - r)