Math 521 Archives - Page 3 of 8 - Shawn Zhong

Math 521 – 4/11

$Theorem 3.55 • Statement ○ If Σa_n is a series of complex numbers which converges absolutely ○ Then every rearrangement of Σa_n converges to the same sum • Proof ○ Let Σa_n^′ be a rearrangement of Σa_n with partial sum s_n^′ ○ By the Cauchy Criterion, given ε0, ∃N∈N s.t. § ∑_(i=n)^m▒|a_i | ε,∀m,n≥N ○ Choose p s.t. 1,2,…,N are all contained in the set {k_1,k_2,…,k_p } ○ Where k_1,…,k_p are the indices of the rearranged series ○ Then if np, a_1,…,a_N will be cancelled in the difference s_n−s_n^′ ○ So, |s_n−s_n^′ |≤ε⇒{s_n^′ } converges to the same value as {s_n } Limit of Functions • Definition ○ Let X,Y be metric spaces, and E⊂X ○ Suppose f:E→Y and p is a limit point of E ○ We write § f(x)→q as x→p, or § (lim)_(x→p)⁡f(x)=q ○ If ∃q∈Y s.t. § Given ε0, there exists δ0 s.t. § If 0d_X (x,p)δ, then d_Y (f(x),q)ε • Note ○ 0d_X (x,p)δ is the deleted neighborhood about p of radius δ ○ d_X and d_Y refer to the distances in X and Y, respectively • Relationship with sequence ○ Theorem 4.2 relates this type of limit to the limit of a sequence ○ Consequently, if f has a limit at p, then its limit is unique Theorem 4.3 • If f,g are complex function on E, we have • (f+g)(x)=f(x)+g(x) • (f−g)(x)=f(x)−g(x) • (fg)(x)=f(x)g(x) • (f/g)(x)=f(x)/g(x) where g(x)≠0 on E Theorem 4.4 (Algebraic Limit Theorem) • Let X be a metric space, E⊂X • Suppose p be a limit point of E • Let f,g be complex functions on E where ○ lim_(x→p)⁡f(x)=A ○ lim_(x→p)⁡g(x)=B • Then ○ lim_(x→p)⁡(f(x)+g(x))=A+B ○ lim_(x→p)⁡(f(x)g(x))=AB ○ lim_(x→p)⁡(f(x)/g(x) )=A/B where B≠0$

Math 521 – 4/9

$Power Series • Given a sequence {c_n } of complex numbers • The series ∑_(n=1)^∞▒〖c_n z^n 〗 is a power series Theorem 3.39 (Convergence of Power Series) • Statement ○ Given the power sires ∑_(n=1)^∞▒〖c_n z^n 〗 ○ Put α≔(lim⁡sup)_(n→∞)⁡√(n&|c_n | ) ○ Let R≔1/α (If α=+∞,R=0; If α=0,R=+∞) ○ Then ∑_(n=1)^∞▒〖c_n z^n 〗 converges if |z| R and diverges if |z| R • Proof ○ Let a_n=c_n z^n and apply the root test ○ (lim⁡sup)_(n→∞)⁡√(n&|a_n | )=|z| (lim⁡sup)_(n→∞)⁡√(n&|c_n | )=|z|/R • Note: R is called the radius of convergence of the power series • Examples ○ ∑_(n=1)^∞▒〖n^n z^n 〗 has R=0 ○ ∑_(n=0)^∞▒z^n/n! has R=+∞ ○ ∑_(n=0)^∞▒z^n has R=1. If |z|=1, then the series diverges ○ ∑_(n=1)^∞▒z^n/n has R=1,diverges if z=1, converges for all other z with |z|=1 ○ ∑_(n=1)^∞▒z^n/n^z has R=1, but converges for all z with |z|=1 by comparison Theorem 3.43 (Alternating Series Test) • Statement ○ Suppose we have a real sequence {c_n } s.t. ○ |c_1 |≥|c_2 |≥|c_3 |≥… ○ c_(2m−1)≥0, c_2m≤0, ∀m∈N ○ lim_(n→∞)⁡〖c_n 〗=0 ○ Then ∑_(n=1)^∞▒c_n converges • Proof: HW • Example: alternating harmonic series ○ ∑_(n=1)^∞▒(−1)^(n+1)/n=1−1/2+1/3−1/4+1/5⋯converges to ln⁡2 Absolute Convergence • The series Σa_n is said to converge absolutely if the series Σ|a_n | converges • If Σa_n converges but Σ|a_n | diverges • We way that Σa_n converges nonabsolutely or conditionally Theorem 3.45 (Property of Absolute Convergence) • Statement ○ If Σa_n converges absolutely, then Σa_n converges • Proof ○ |∑_(k=1)^∞▒a_k |≤∑_(n=k)^∞▒|a_k | ○ The result follows by Cauchy Criterion Rearrangement • Let {k_n } be a sequence in which every natural number appears exactly once • Let a_n^′=a_(k_n ), then Σa_n^′ is called a rearrangement of Σa_n Theorem 3.54 (Riemann Series Theorem) • Let Σa_n be a series of real number which converges nonabsolutely • Let −∞≤α≤β≤+∞ • Then there exists a rearrangement Σa_n^′ s.t. • (lim⁡inf)_(n→∞)⁡〖s_n^′ 〗=α, (lim⁡sup)_(n→∞)⁡〖s_n^′ 〗=β Theorem 3.55 • If Σa_n is a series of complex numbers which converges absolutely • Then every rearrangement of Σa_n converges to the same sum$

Math 521 – 4/6

$Theorem 3.27 (Cauchy Condensation Test) • Statement ○ Suppose a_1≥a_2≥…≥0, then ○ ∑_(n=1)^∞▒a_n converges⟺∑_(k=0)^∞▒〖2^k a_(2^k ) 〗=a_1+2a_2+4a_4+…converges • Proof ○ By Theorem 3.24, we just need to look at boundness of partial sums ○ Let § s_n=a_1+a_2+…+a_n, § t_k=a_1+2a_2+…+2^k a_(2^k ) ○ For n≤2^k § s_n≤a_1+(a_2+a_3 )+…+(a_(2^k )+…+a_(2^(k+1)−1) ) § ≤a_1+2a_2+…+2^k a_(2^k )=t^k ○ For n≥2^k § s_n≥a_1+(a_2+a_3 )+…+(a_(2^(k−1)+1)+…+a_(2^k ) ) § ≥1/2 a_1+a_2+…+2^(k−1) a_(2^k )=1/2 t^k ○ For n=2^k § s_n≤t_k≤2s_n⇒s_(2^k )≤t_k≤2s_(2^k ) § So {s_n } and {t_k } are both bounded or unbounded Theorem 3.28 • Statement ○ ∑_(n=1)^∞▒1/n^p converges if p 1 and diverges if p≤1 • Proof ○ If p≤0 § Theorem 3.23 says if∑_(n=1)^∞▒a_n converges, then lim_(n→∞)⁡〖a_n 〗=0 § In this case lim_(n→∞)⁡〖1/n^p 〗≠0, so series diverges ○ If p 0 § 1/n^p ≥1/(n+1)^p and 1/n^p ≥0 § By Cauchy Condensation Test, § lim_(n→∞)⁡〖1/n^p 〗 converges⟺∑_(n=1)^∞▒〖2^k 1/(2^k )^p 〗 converges § ∑_(n=1)^∞▒〖2^k 1/(2^k )^p 〗=∑_(n=1)^∞▒(2^(1−p) )^k which is a geometric series § By Theorem 3.26, this converges if 2^(1−p) 1⟺p 1 § Otherwise, 2^(1−p) 1, and this diverges Theorem 3.33 (Root Test) • Given ∑_(n=1)^∞▒a_n , put α=(lim⁡sup)_(n→∞)⁡√(n&|a_n | ), then • If α 1, ∑_(n=1)^∞▒a_n converges ○ Theorem 3.17(b) says if x s^∗,then ∃N∈N s.t.s_n x for n≥N ○ So let β∈(α,1) and N∈N s.t. ∀n≥N, √(n&|a_n | ) β i.e. |a_n | β^n ○ 0 β 1, so ∑_(n=1)^∞▒β^n converges ○ Thus, ∑_(n=1)^∞▒a_n converges by comparison test • If α 1, ∑_(n=1)^∞▒a_n diverges ○ By Theorem 3.17, there exists a sequence {n_k } s.t. √(n_k&|a_(n_k ) | )→α ○ So |a_n | 1 for infinitely many n, i.e. a_n↛0 ○ By Theorem 3.23, ∑_(n=1)^∞▒a_n diverges • If α=1, this test gives no information ○ For ∑_(n=1)^∞▒1/n, (lim⁡sup)_(n→∞)⁡√(n&n^(−1) )=lim_(n→∞)⁡√(n&n^(−1) )=1, but the series diverges ○ For ∑_(n=1)^∞▒1/n^2 , (lim⁡sup)_(n→∞)⁡√(n&n^(−2) )=lim_(n→∞)⁡〖1/(√(n&n))^2 〗=1, but the series converges Theorem 3.34 (Ratio Test) • Statement ○ ∑_(n=1)^∞▒a_n converges if (lim⁡sup)_(n→∞)⁡〖|a_(n+1)/a_n | 1〗 ○ ∑_(n=1)^∞▒a_n diverges if |a_(n+1)/a_n |≥1,∀n≥n_0 for some fixed n_0∈N • Proof ○ If (lim⁡sup)_(n→∞)⁡〖|a_(n+1)/a_n | 1〗 ○ We can find β 1, N∈N s.t. |a_(n+1)/a_n | β,∀n≥N ○ In particular § |a_(N+1) | β|a_N | § |a_(N+2) | β|a_(N+1) | β^2 |a_N | § ⋮ § |a_(N+p) | β^p |a_N | ○ So, |a_n | |a_N | β^(−N) β^n, ∀n≥N ○ β 1, so ∑_(n=1)^∞▒β^n converges ○ So ∑_(n=1)^∞▒〖⏟(|a_N | β^(−N) )┬constant β^n 〗 also converges ○ Therefore ∑_(n=1)^∞▒a_n converges by comparison test ○ On the other hand, if |a_(n+1) |≥|a_n |,∀n≥n_0∈N ○ Then a_n↛0, so series divreges by Theorem 3.23 • Note ○ For ∑_(n=1)^∞▒1/n, lim_(n→∞)⁡〖(1/n)/(1/(n+1) )〗=1 ○ For ∑_(n=1)^∞▒1/n^2 , lim_(n→∞)⁡〖(1/n^2)/(1/(n+1)^2 )〗=1 ○ So lim_(n→∞)⁡〖a_n/a_(n+1) 〗=1 is not enough to conclude anything$

Math 521 – 4/4

$Series • Given a sequence {a_n } • We associate a sequence of partial sums {s_n } where • s_n=∑_(k=1)^n▒a_k =a_1+a_2+…+a_n • ∑_(k=1)^∞▒a_k is called an infinite series, or simply series • If {s_n } diverges, the series is said to diverge • If {s_n } converges to s, the series is said to converge, and write ∑_(k=1)^∞▒a_k =s • s is called the sum of the series • But it is technically the limit of a sequence of sums Theorem 3.22 (Cauchy Criterion for Series) • Statement ○ ∑_(n=1)^∞▒a_n converges⟺∀ε 0, ∃N∈N s.t. |∑_(k=n)^m▒a_k | ε,∀m≥n≥N • Proof ○ This is Theorem 3.11 applied to {s_n } Theorem 3.23 • Statement ○ In the setting of Theorem 3.22, take m=n ○ We have |a_n | ε for n≥N ○ If ∑_(n=1)^∞▒a_n converges, then (lim)_(n→∞)⁡〖a_n 〗=0 • Note ○ If a_n→0, the series ∑_(n=1)^∞▒a_n might not converge • Example: ∑_(n=1)^∞▒1/n diverges ○ 1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+…≥1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+… ○ Therefore ∑_(n=1)^∞▒1/n diverges Theorem 3.24 • Statement ○ A series of nonnegative real numbers converges if and only if ○ its partial sum form a bounded sequence • Proof ○ See Theorem 3.14 (Monotone Convergence Theorem) Theorem 3.25 (Comparison Test) • If |a_n | c_n for n≥N_0∈N and ∑_(n=1)^∞▒c_n converges, then ∑_(n=1)^∞▒a_n converges ○ Given ε 0,∃N≥N_0 s.t. |∑_(k=n)^m▒c_k |=∑_(k=n)^m▒c_k ε for m≥n≥N ○ By the Cauchy Criterion, |∑_(k=n)^m▒a_k |≤∑_(k=n)^m▒|a_k | ≤∑_(k=n)^m▒c_k ε ○ Thus ∑_(n=1)^∞▒a_n converges • If a_n≥d_n≥0 for n≥N_0∈N and ∑_(n=1)^∞▒d_n diverges, then ∑_(n=1)^∞▒a_n diverges ○ If ∑_(n=1)^∞▒a_n converges, then so must ∑_(n=1)^∞▒d_n ○ This is a contradiction, so ∑_(n=1)^∞▒a_n diverges Theorem 3.26 • Statement ○ If 0 x 1, then∑_(n=0)^∞▒x^n =1/(1−x) ○ If x 1, the series diverges • Note ○ {█(S=1+x+x^2+…@xS=x+x^2+…)┤⇒S−xS=1⇒S=1/(1−x) ○ This only works if we know this series converges • Proof ○ If 0 x 1, we have ○ {█(s_n=1+x+x^2+…+x^n@xs_n=x+x^2+…+x^n+x^(n+1) )┤ ○ ⇒s_n−xs_n=1−x^(n+1) ○ ⇒s_n=(1−x^(n+1))/(1−x) ○ Since 0 x 1,lim_(n→∞)⁡〖s_n 〗=lim_(n→∞)⁡〖(1−x^(n+1))/(1−x)〗=1/(1−x) ○ Note if x=1, ∑_(n=1)^∞▒x^n =1+1+…which diverges$

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)⇒n^p 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≥1^n+(█(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)/(2^k 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=(2^k 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

Shawn Zhong

Shawn Zhong

Math 521

Math 521 – 4/11

Math 521 – 4/9

Math 521 – 4/6

Math 521 – 4/4

Math 521 – 4/2

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