Math 521 - 3/5

Theorem 3.3 (Algebraic Limit Theorem) • Suppose {s_n },{t_n } are complex sequence, and lim_(n→∞)⁡〖s_n 〗=s,lim_(n→∞)⁡〖t_n 〗=t, then • (lim)_(n→∞)⁡〖s_n+t_n 〗=s+t ○ Given ε 0 § lim_(n→∞)⁡〖s_n 〗=s⇒∃N_1∈N s.t. |s_n−s| ε/2 for n≥N_1 § lim_(n→∞)⁡〖t_n 〗=t⇒∃N_2∈N s.t. |t_n−t| ε/2 for n≥N_2 ○ Let N=max⁡(N_1,N_2 ), then for n≥N § |s_n+t_n−(s+t)|=|(s_n−s)+(t_n−t)|≤|s_n−s|+|t_n−t| ε ○ Therefore lim_(n→∞)⁡〖s_n+t_n 〗=s+t • (lim)_(n→∞)⁡〖c+s_n 〗=c+s, ∀c∈ℂ ○ Given ε 0 ○ lim_(n→∞)⁡〖s_n 〗=s⇒∃N∈N s.t.|s_n−s| ε for n≥N ○ So, |c+s_n−(c+s)|=|s_n−s| ε ○ Therefore lim_(n→∞)⁡〖c+s_n 〗=c+s • (lim)_(n→∞)⁡〖cs_n 〗=cs, ∀c∈ℂ ○ Given ε 0 ○ If c=0 § |cs_n−cs|=0 ε ○ If c≠0 § lim_(n→∞)⁡〖s_n 〗=s⇒∃N∈N s.t. |s_n−s| ε/|c| for n≥N § So |cs_n−cs|=|c||s_n−s| |c| ε/|c| =ε ○ Therefore lim_(n→∞)⁡〖cs_n 〗=cs • (lim)_(n→∞)⁡〖s_n t_n 〗=st ○ Standard approach § s_n t_n−st=s_n t_n−st_n+st_n−st=t_n (s_n−s)+s(t_n−t) ○ Rudin s approach § s_n t_n−st=(s_n−s)(t_n−t)+t(s_n−s)+s(t_n−t) ○ Given ε 0 § ∃N_1∈N s.t. |s_n−s| √ε for n≥N_1 § ∃N_2∈N s.t. |t_n−t| √ε for n≥N_2 ○ Let N=max⁡(N_1,N_2 ),then § |(s_n−s)(t_n−t)| ε for n≥N § ⇒lim_(n→∞)⁡(s_n−s)(t_n−t)=0 ○ lim_(n→∞)⁡〖s_n t_n 〗 § =lim_(n→∞)⁡[(s_n−s)(t_n−t)+t(s_n−s)+s(t_n−t)+st] § =lim_(n→∞)⁡(s_n−s)(t_n−t)+t lim_(n→∞)⁡(s_n−s)+s lim_(n→∞)⁡(t_n−t)+st § =0+0+0+st § =st ○ Therefore lim_(n→∞)⁡〖s_n t_n 〗=st • (lim)_(n→∞)⁡〖1/s_n 〗=1/s (s_n≠0, ∀n∈N and s≠0) ○ lim_(n→∞)⁡〖s_n 〗=s⇒∃N^′∈N s.t. |s_n−s| |s|/2 for n≥N^′ ○ By the Triangle Inequality, |s|−|s_n |≤|s_n−s|,∀n≥N^′ ○ ⇒|s_n |≥|s|−|s_n−s| |s|−|s|/2=|s|/2,∀n≥N^′ ○ Given ε 0, ∃N N^′ s.t. |s_n−s| 1/2 |s|^2 ε for n≥N ○ |1/s_n −1/s|=|(s−s_n)/(s_n s)| (1/2 |s|^2 ε)/(|s_n |⋅|s| ) (1/2 |s|^2 ε)/(|s|/2⋅|s| )=ε ○ Therefore lim_(n→∞)⁡〖1/s_n 〗=1/s