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Power Series

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Definitions of Power Series
Power Series
∑n=0∞cn(x−a)n=c0+c1(x−a)+c2(x−a)2+c3(x−a)3+...\sum\limits^\infty_{n=0}c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3+...n=0∑∞​cn​(x−a)n=c0​+c1​(x−a)+c2​(x−a)2+c3​(x−a)3+...
where xxx is a variable,aaa is a constant, and the terms cnc_ncn​ are coefficients. 
*We say that this is a power series "centered at x=ax=ax=a" or "about x=ax=ax=a"
Note: unlike a normal series that only have nnn, power series depend on the variable xxx.

Special Case of Power Series
If the power series is centered at x=0x=0x=0
 ∑n=0∞cnxn=c0+c1x+c2x2+c3x3+...\sum\limits^\infty_{n=0}c_nx^n=c_0+c_1x+c_2x^2+c_3x^3+...n=0∑∞​cn​xn=c0​+c1​x+c2​x2+c3​x3+...

Radius of Convergence
The convergence of the power series might depend on the value of xxx.
The radius of convergence is denoted by a positive value RRR, and it helps give a range/interval for xxx on which the power series converges.

There are three cases for a power series about x=ax=ax=a: 
1. The series converges if ∣x∣|x|∣x∣ < RRR but diverges if ∣x∣|x|∣x∣ > RRR. 
    *Note that when x=±Rx=\pm Rx=±R, whether the series converge depends on the series itself
    *So, the possible intervals of convergence are (a-R, a+R), [a-R, a+R], (a-R, a+R] or [a-R, a+R)

2. The series converged for all values of xxx (i.e. R=∞R=\inftyR=∞)

3. The series only converges for xxx = aaa (i.e. R=0R=0R=0)

How to find the Radius/Interval of Convergence?
1. Use the Ratio test or Root test on cn(x−a)nc_n(x-a)^ncn​(x−a)n to determine the radius of convergence RRR

2. If RRR is a finite number (≠0 or ∞\ne0\ or\ \infty=0 or ∞) , we will have to check the convergence on both endpoints x=R,  x=−Rx=R,\ \ x=-Rx=R,  x=−R to see if we should include them in the interval of convergence.
*Use convergence tests from the previous chapter to see if the series converges at these point

Practice: Radius of Convergence
Determine the radius of convergence of the power series ∑n=0∞n! xn{\displaystyle \sum_{n=0}^\infty}n!\ x^nn=0∑∞​n! xn.

Practice: Radius & Interval of Convergence
Determine the radius and interval of convergence of the power series ∑n=0∞(−2)nn+1(x+1)n{\displaystyle \sum_{n=0}^\infty}
\frac{\left(-2\right)^n}{\sqrt{n+1}}\left(x+1\right)^nn=0∑∞​n+1​(−2)n​(x+1)n.

R=12R=\frac{1}{2}R=21​
Interval of convergence is [−32,−12]\left[-\frac{3}{2},-\frac{1}{2}\right][−23​,−21​]

R=12R=\frac{1}{2}R=21​
Interval of convergence is (−32,−12]\left(-\frac{3}{2},-\frac{1}{2}\right](−23​,−21​]

R=12R=\frac{1}{2}R=21​
Interval of convergence is [−32,−12)\left[-\frac{3}{2},-\frac{1}{2}\right)[−23​,−21​)

R=1R=1R=1
Interval of convergence is [−32,−12]\left[-\frac{3}{2},-\frac{1}{2}\right][−23​,−21​]

R=1R=1R=1
Interval of convergence is (−32,12)\left(-\frac{3}{2},\frac{1}{2}\right)(−23​,21​)

I don't know

Practice: Radius & Interval of Convergence
Determine the radius and interval of convergence of the power series ∑n=1∞(−1)n3nnxn{\displaystyle \sum_{n=1}^\infty}
\frac{\left(-1\right)^n}{3^nn}x^nn=1∑∞​3nn(−1)n​xn.

R=3R=3R=3
Interval of convergence is (−3,3)\left(-3,3\right)(−3,3)

R=3R=3R=3
Interval of convergence is (−3,3]\left(-3,3\right](−3,3]

R=3R=3R=3
Interval of convergence is [0,3)\left[0,3\right)[0,3)

R=1R=1R=1
Interval of convergence is (−1,1)\left(-1,1\right)(−1,1)

R=1R=1R=1
Interval of convergence is [−1,1]\left[-1,1\right][−1,1]

I don't know