- #1
synkk
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Given a sequence ## <x_n> ##, let ## <x_{n+1}> ## denote the sequence whose nth term for each ## n \in \mathbb{N} ## is ## x_{n+1} ##. Show that if ## <x_n> ## converges then ## < x_{n+1} ## converges and they have the same limit.
my attempt thus far
given ## \epsilon > 0 ## ##\exists N \in \mathbb{R} ## s.t. n > N implies ## |x_n - l | < \epsilon ## where l is the limit of ## x_n##.
now we need to prove that for given ## \epsilon > 0 ## ##\exists N_1 \in \mathbb{R} ## s.t. ##n > N_1 ## implies ## |x_{n+1} - m | < \epsilon ## where m is the limit of ## x_{n+1} ## and m = l
heres what I have thus far
consider ## |x_n - l |## :
## |x_n - l | = |x_n - l + x_{n+1} - x_{n+1} | = |(x_n - x_{n+1}) + (x_{n+1} - l ) | \leq |x_n - x_{n+1} | + |x_{n+1} - l| \leq |x - x_{n+1} | + |x_{n+1} - l | \leq |x_{n+1} - x | + |x_{n+1} -l | ##
any ideas how to proceed? or if I'm even doing it correctly
my attempt thus far
given ## \epsilon > 0 ## ##\exists N \in \mathbb{R} ## s.t. n > N implies ## |x_n - l | < \epsilon ## where l is the limit of ## x_n##.
now we need to prove that for given ## \epsilon > 0 ## ##\exists N_1 \in \mathbb{R} ## s.t. ##n > N_1 ## implies ## |x_{n+1} - m | < \epsilon ## where m is the limit of ## x_{n+1} ## and m = l
heres what I have thus far
consider ## |x_n - l |## :
## |x_n - l | = |x_n - l + x_{n+1} - x_{n+1} | = |(x_n - x_{n+1}) + (x_{n+1} - l ) | \leq |x_n - x_{n+1} | + |x_{n+1} - l| \leq |x - x_{n+1} | + |x_{n+1} - l | \leq |x_{n+1} - x | + |x_{n+1} -l | ##
any ideas how to proceed? or if I'm even doing it correctly