Find the following power series

Please help me find the sigma notation for any of the following expressions:

x/(1-x)

2*(cos(x^2))

ln(1-x)

1/(1+x)^2


Thank you!!

Assuming these series are to be centered at x = 0:

1) Starting with the geometric series 1/(1 - x) = Σ(n = 0 to ∞) x^n:

Multiply both sides by x:
x/(1 - x) = Σ(n = 0 to ∞) x^(n+1).

2) Starting with cos t = Σ(n = 0 to ∞) (-1)^n t^(2n)/(2n)!:

Let t = x^2:
cos(x^2) = Σ(n = 0 to ∞) (-1)^n x^(4n)/(2n)!

Multiply both sides by 2:
2 cos(x^2) = Σ(n = 0 to ∞) 2 (-1)^n x^(4n)/(2n)!

3) Starting with the geometric series 1/(1 - x) = Σ(n = 0 to ∞) x^n:

Integrate both sides from 0 to x:
-ln(1 - x) = Σ(n = 0 to ∞) x^(n+1)/(n+1)
==> ln(1 - x) = -Σ(n = 0 to ∞) x^(n+1)/(n+1)

4) Starting with the geometric series 1/(1 - x) = Σ(n = 0 to ∞) x^n:

Differentiate both sides:
1/(1 - x)^2 = Σ(n = 1 to ∞) nx^(n-1)

Replace x with -x:
1/(1 + x)^2 = Σ(n = 1 to ∞) n(-1)^(n-1) x^(n-1).

I hope this helps!

First recall : 1/(1-x) = 1 + x + x^2 + x^3 +....
Now,
x/(1-x) = x( 1 + x + x^2 + x^3 +....) = x + x^2 + x^3 +.....
ln(1-x) = -integral(1/(1-x)) = -(x + (x^2)/2 +(x^3)/3 +.....)
1/(1+x)^2 = -(1/(1+x))' = -(1 -x + x^2 - x^3+...)' = 1 -2x + 3x^2 -4x^3+....

Recall : cos(x) = 1 -(x^2)/2! + (x^4)/4! - (x^6)/6! +......
So, 2(cos(x^2)) = 2( 1 -(x^4)/2! + (x^6)/4! - (x^8)/6! +......) = 2 - 2(x^4)/2! + 2 (x^6)/4 - 2(x^8)/6! +...