A series of the form:
∑ n = 0 ∞ a n ( x − c ) n \sum_{n=0}^\infty
a_n (x-c)^n
n = 0 ∑ ∞ a n ( x − c ) n Here:
x x x c c c
Convergence of a power series can be checked using
ratio test or
root test .
Maximum radius of x x x
R = sup { r ∣ series converges for ∣ x − c ∣ < r } R = \sup{\big\{r\;|\; \text{series converges for}\; \lvert x-c \rvert < r\big\}}
R = sup { r ∣ series converges for ∣ x − c ∣ < r } The below equation can be used to find R R R
lim k → ∞ ∣ a k ∣ 1 k = 1 R \lim_{k\to \infty} |a_k|^{\frac{1}{k}} = \frac{1}{R}
k → ∞ lim ∣ a k ∣ k 1 = R 1 The series may converge or diverge for ∣ x − c ∣ = R \lvert x - c \rvert = R ∣ x − c ∣ = R
( c − R , c + R ) (c-R,c+R) ( c − R , c + R )
The series is:
Absolutely converging for ∣ x − a ∣ < R \lvert x-a \rvert \lt R ∣ x − a ∣ < R
Diverging for ∣ x − a ∣ < R \lvert x-a \rvert \lt R ∣ x − a ∣ < R
Uniformly converging for ∣ x − a ∣ ≤ ρ < R \lvert x-a \rvert \le \rho \lt R ∣ x − a ∣ ≤ ρ < R
Suppose R ∈ ( 0 , ∞ ) R \in (0,\infty) R ∈ ( 0 , ∞ ) 0 < p < R 0 \lt p \lt R 0 < p < R ∀ x ∀ n \forall x \forall n ∀ x ∀ n ∣ x − a ∣ ≤ p ⟹ s n ( x ) ) \lvert x-a \rvert \le p \implies s_n(x)) ∣ x − a ∣ ≤ p ⟹ s n ( x ))
Note the relation between R R R a k a_k a k
Prove ( p + R 2 p R ) k (\frac{p+R}{2pR})^k ( 2 pR p + R ) k ∣ a k ∣ 1 k |a_k|^{\frac{1}{k}} ∣ a k ∣ k 1
Define M k = ( p + R 2 R ) k = r k M_k = (\frac{p+R}{2R})^k = r^k M k = ( 2 R p + R ) k = r k
Prove M k M_k M k u k u_k u k
Prove ∑ k = 1 n r k \sum_{k=1}^{n} r^k ∑ k = 1 n r k 0 < r < 1 0\lt r\lt 1 0 < r < 1