Infinate series

[scott73]

I cannot figure this problem out at all, I cant tell what type of series it is and at what point it converges too, I do know that it converges though,
here it is, any help would be awesome!

(The Summation of, from n=1 to infinity) ((-2)^(n-1))/(8^n)

 

[stapel]

I'm not sure what you mean by "what type of series it is", other than "infinite". As for working with the series terms, you might want to note that (-2)^n-1 = (-1)^n-1 (2)^n-1, and 8ⁿ = (2³)ⁿ = 2³ⁿ.

Eliz.

 

[galactus]

This is a cool infinite series. It does converge.

Let's step through it:

 

[soroban]

Hello, scott73!

_oo . (-2)^n-1
∑ . ---------
ⁿ⁼¹ . 8ⁿ


The numerator is: .(-1)^n-1(2^n-1)

The denominator is: .8ⁿ .= .(2³)ⁿ .= .2³ⁿ

. . . . . . . . . . . . . .(-1)^n-1(2^n-1) . . . . (-1)^n-1
The function is: . ---------------- . = . ----------
. . . . . . . . . . . . . . . . . 2³ⁿ . . . . . . . . .2²ⁿ⁺¹


The series is: . 1/2³ - 1/2⁵ + 1/2⁷ - 1/2⁹ + . . .

This is a geometric series with first term a = 1/8 and common ratio r = -1/4

. . . . . . . . . . .1 . . . . . 1 . . . . . . . 1
The sum is: . -- . ------------ . = . --- . . . . . . exactly what galactus predicted!
. . . . . . . . . . .8. . 1 - (-1/4) . . . .10