Math 521 - 4/2

Theorem 3.20 • Lemma (The Squeeze Theorem) ○ Given 0≤x_n≤s_n, for n≥N where N∈N is some fixed number ○ If s_n→0, then x_n→0 ○ (Proof on homework) • If p 0, then (lim)_(n→∞)⁡〖1/n^p 〗=0 ○ For n≥N, we need |1/n^p −0| ε⇒n 1/ε^(1∕p) ○ Given ε 0 ○ Using Archimedean Property, take N (1/ε)^(1/p) ○ So, for n≥N, n (1/ε)^(1/p)⇒np 1/ε⇒1/n^p ε⇒|1/n^p −0| ε ○ Therefore lim_(n→∞)⁡〖1/n^p 〗=0 • If p 0, then (lim)_(n→∞)⁡√(n&p)=1 ○ When p=1 § We are done, since lim_(n→∞)⁡1=1 ○ When p 1 § Then p−1 0 § Let x_n=√(n&p)−1, then x_n 0 § p=(x_n+1)^n≥1n+(█(n@n−1)) 1^(n−1) x_n^1=1+nx_n § ⇒p−1≥nx_n § ⇒(p−1)/n≥x_n 0 § By the Squeeze Theorem, x_n→0 § i.e.lim_(n→∞)⁡〖√(n&p)−1〗=0 § So lim_(n→∞)⁡√(n&p)=1 ○ When p 1 § Then 1/p 1 § So, lim_(n→∞)⁡√(n&1∕p)=1 § Therefore lim_(n→∞)⁡√(n&p)=1/1=1 • (lim)_(n→∞)⁡√(n&n)=1 ○ Let x_n=√(n&n)−1≥0 ○ n=(x_n+1)^n≥(█(n@n−2)) 1^(n−2) x_n^2=n!/(n−2)!2! x_n^2=n(n−1)/2 x_n^2 ○ ⇒2/(n−1)≥x_n^2 ○ ⇒√(2/(n−1))≥x_n 0 for n 1 ○ By the Squeeze Theorem, x_n=lim_(n→∞)⁡〖√(n&n)−1〗→0 ○ i.e. lim_(n→∞)⁡√(n&n)=1 • If p 0,α∈R, then (lim)_(n→∞)⁡〖n^α/(1+p)n 〗=0 ○ Let k∈N s.t. k α by Archimedean Property ○ For n 2k,(1+p)^n (█(n@k)) p^k=(n(n−1)⋯(n−k+1))/k! p^k (n^k p^k)/(2k k!) ○ Because n 2k⇒n/2 k⇒n−k n/2⇒n−k+1 n/2 ○ So, 0 n^α/(1+p)α (2^k k!)/(n^k p^k )⋅n^α=(2k k!)/p^k ⋅n^(α−k) ○ Since a−k 0, n^{(a−k)→0⇒(2}k k!)/p^k ⋅n^(α−k)→0 ○ By the Squeeze Theorem, n^α/(1+p)α →0 ○ i.e. lim_(n→∞)⁡〖n^α/(1+p)n 〗=0 • If |x| 1, then (lim)_(n→∞)⁡〖x^n 〗=0 ○ |x| 1⇒1/|x| 1 ○ Let p=1/|x| −1 0 ○ Take α=0 in the limit above, we get lim_(n→∞)⁡〖1/(1+p)^n 〗=0 ○ So lim_(n→∞)⁡〖1/((1+1/|x| −1)^n )〗=lim_(n→∞)⁡〖|x|^n 〗=0 ○ Then lim_(n→∞)⁡〖x^n 〗=0