Question

\[\sum_{n=1}^{\infty} 2(\frac{ 9 }{ 10 })^{n}\] is there a sum to the series?

Answer 1

there is in fact a sum to the series because r= 9/10 and IrI<1

Answer 2

true story

Answer 3

so it converges to 2?

Answer 4

\[\sum_{0}^{\infty} r^{n}\] converges to \[\frac{ 1 }{ 1- r }\] if r < 1

we are dealing with \[2\sum_{0}^{\infty} (9/10)^{n}\ = 2*\frac{ 1 }{ 1 - (9/10) }\]

Answer 5

so it would converge to 2/.10

Answer 6

isnt it 1 to infinity, not 0 to infinity?

Answer 7

you're right blur

Answer 8

oh it's when n=0

Answer 9

and because n can get really big and anything to the ^1 = 1, the sum will be 2?

Answer 10

my mistake; good observation blur.

\[2\sum_{0}^{\infty - 1}(9/10)^{n-1} = 2\sum_{0}^{\infty}(9/10)^{n}*\frac{ 1 }{ (9/10) }\]

\[= (20/9)\sum_{0}^{\infty}(9/10)^{n}\]

rule applies

Answer 11

So the sum of this series is 20?

Answer 12

no it's: \[(20/9)*\frac{ 1 }{ 1 - (9/10) }\] . doens't look like it'll be an integer

Answer 13

2.2 x 1.1 = 2.42?

Answer 14

1/(1-9/10) doesn't = 1.1

Answer 15

ok, thank you