Find Power Series?

Question

Find Power Series and determine the Interval of Convergence

g(x)=3x/(x^2+x-2) , c=0
- after doing partial fractions and getting the form into a/1-r and finding endpoints, then combining both and I got the Interval of Convergence to be (-1,1) is that right?

f(x)=1/2-x ,c=5
- I tried this 1 way but the book did it another way and got a diff. answer so can someone work this ?

Use the Power Series 1/1+x = E (-1)^n*x^n to determine a power series , c=0. Identify Interval of Converg.

f(x)=1/(x+1)^2 = d/dx [1/x+1]
- I dont get how to do this

f(x)=ln(x+1) = S 1/x+1 dx
- this one either

Answer 1

g(x) = 3x / (x^2 + x - 2), c = 0
Partial fractions -> g(x) = 2 / (x+2) + 1 / (x-1)
For the power series I wind up with Σ(n=0 to ∞) x^n ((-1/2)^n - 1).
Ratio of n+1'th to n'th term is x ((-1/2)^(n+1) - 1) / ((-1/2)^n - 1)
and (1/2)^n -> 0 as n -> ∞, so in the limit the absolute value of the ratio goes to |x|. So radius of convergence is 1.
Check the endpoints: for x = 1 we get Σ(n=0 to ∞) ((-1/2)^n - 1) and for x = -1 we get Σ(n=0 to ∞) (-1)^n ((-1/2)^n - 1). In the first case the terms converge to -1 and in the second they alternate between +1 and -1; since in both cases the terms do not go to 0 the series cannot converge. So the interval of convergence is (-1, 1).

f(x) = 1 / (2-x), c = 5; let y = x - 5
f(x) = g(y) = 1 / (-3-y) = (-1/3) . 1 / (1 + y/3)
= (-1/3) Σ(n=0 to ∞) (-y/3)^n
= (-1/3) Σ(n=0 to ∞) ((5-x)/3)^n
Absolute value of the ratio of successive terms is |5-x| / 3 so the radius of convergence is 3 and the interval of convergence has interior (2, 8). Check the endpoints:
x = 2: (-1/3) Σ(n=0 to ∞) 1 diverges
x = 8: (-1/3) Σ(n=0 to ∞) (-1)^n diverges
So the interval of convergence is (2, 8).

For the last two, the point is that if you know the power series for the derivative or the integral of the function, you can obtain the power series of the function itself by integrating or differentiating term by term, respectively.

Minor error on the first one - a negative sign is missing.
f(x) = 1/(x+1)^2 = d/dx(-1/(x+1))
= d/dx (Σ(n=0 to ∞) (-1)^(n+1) x^n)
= Σ(n=1 to ∞) (-1)^(n+1) (nx^(n-1))
(note I've dropped n=0 because the derivative of that term is 0).
Abs. ratio is |(n+1) x / n| -> |x| as n -> ∞. So the radius of convergence is 1. Check the endpoints:
x = 1 gives Σ(n=1 to ∞) (-1)^(n+1) (n)
x = -1 gives Σ(n=1 to ∞) n
which both obviously diverge. So the interval of convergence is (-1, 1).

f(x) = ln (x+1) = ∫1/(x+1) dx
= ∫(Σ(n=0 to ∞) (-1)^n x^n) dx
= Σ(n=0 to ∞) (-1)^n x^(n+1) / (n+1) + k
when x = 0, we get 0 = k, so the power series is
Σ(n=0 to ∞) (-1)^n x^(n+1) / (n+1)
which has abs. ratio |x (n+2) / (n+1)| -> |x| as n -> ∞.
So the radius of convergence is 1.
Check the endpoints:
for x = 1 we get Σ(n=0 to ∞) (-1)^n / (n+1)
which is the alternating harmonic series, so converges
and for x = -1 we get Σ(n=0 to ∞) -1 / (n+1)
which is the harmonic series, so diverges.
So the interval of convergence is (-1, 1].