- #1
synkk
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prove if ## a_{2k} \rightarrow l ## and ## a_{2k-1} \rightarrow l ## then ## a_n \rightarrow l ## where ## a_{2k} ## and ## a_{2k-1} ## are subsequences of ## a_n ##
my attempt:
since: ## a_{2k} \rightarrow l ## then ## \forall \epsilon > 0 ## ##\exists N_1 \in \mathbb{R}## s.t. ##2k > N_1 \Rightarrow |a_{2k} - l|< \epsilon##
also, since ## a_{2k-1} \rightarrow l ## then ## \forall \epsilon > 0 ## ##\exists N_2 \in \mathbb{R} ## s.t. ## 2k -1 > N_2 \Rightarrow |a_{2k-1} - l | < \epsilon ##
from the above we have ## k > \dfrac{N_1}{2} ## and ## k > \dfrac{N_2 + 1}{2} ## then let ## N = max \{ \dfrac{N_1}{2}, \dfrac{N_2 + 1}{2} \} ## then for ## n >N ## ## |a_n - l | < \epsilon ##
does this proof make sense, if not where could I improve?
my attempt:
since: ## a_{2k} \rightarrow l ## then ## \forall \epsilon > 0 ## ##\exists N_1 \in \mathbb{R}## s.t. ##2k > N_1 \Rightarrow |a_{2k} - l|< \epsilon##
also, since ## a_{2k-1} \rightarrow l ## then ## \forall \epsilon > 0 ## ##\exists N_2 \in \mathbb{R} ## s.t. ## 2k -1 > N_2 \Rightarrow |a_{2k-1} - l | < \epsilon ##
from the above we have ## k > \dfrac{N_1}{2} ## and ## k > \dfrac{N_2 + 1}{2} ## then let ## N = max \{ \dfrac{N_1}{2}, \dfrac{N_2 + 1}{2} \} ## then for ## n >N ## ## |a_n - l | < \epsilon ##
does this proof make sense, if not where could I improve?