Find the infinite sum of (2/3)+(2/3^3)+(2/3^5)+(2/3^7)+...

Please help, how do you find this sum of the infinite series?

Also 1-(3^2/2!)+(3^3/3!)-(3^4/4!)+(3^5/5!)

(1) Note that this is an infinite geometric series with a first term of 2/3 and a common ratio of 1/3^2 = 1/9.

Using the formula for the sum of an infinite geometric series:
2/3 + 2/3^3 + 2/3^5 + 2/3^7 = (2/3)/(1 - 1/9) = 6/(9 - 1) = 3/4.

(2) The factorials suggest to use the following Maclaurin Series:
sum(n=0 to infinity) x^n/n! = e^x.

If we replace x with -x, we get:
sum(n=0 to infinity) (-x)^n/n! = e^(-x)
==> sum(n=0 to infinity) [(-1)^n * x^n]/n! = 1/e^x.

Letting x = 3 gives the desired series.
sum(n=0 to infinity) [(-1)^n * 3^n]/n! = 1/e^3
==> 1 - 3^2/2! + 3^3/3! - 3^4/4! + 3/5^5! - ... = 1/e^3.

I hope this helps!

Sn = a(1 - r^n)/(1 - r)

=> S∞ = a/(1 - r)...........where a = 2/3 and r = 1/3² = 1/9

=> S∞ = (2/3)/(1 - 1/9) = (2/3)/(8/9) = (2/3) x (9/8) = 18/24 = 3/4

:)>