Proof for a Sequence Convergence

  • Thread starter Bachelier
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  • #1
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[itex] \text{We need to prove that the sequence} \ a_{n} = \{n^{2}/2^{n}\} \ \text{converges to 0} \\

\text{Consider the sequence {n/ 2n} = { 0, 1/2, 1/2, 3/8, 1/4, 5/32, ...}. The terms get smaller and smaller.}\\
\\
\text{we can easily show that} \ n/2^{n}<=1/n \ \forall n>3 \\
\text{from the fact that} \ n^{2}<=2^{n} \ \text{(insert proof by induction here)}\\

then \ for \ \epsilon > 0 \ choose \ N = max\{3,1/\epsilon\} \\

\text{The idea is that for all n>N, we will have} \ 1/n<1/N< \epsilon. \\
\text{The problem is I think N needs to be greater than the max. i.e. } \ N> max\{3,1/\epsilon\} \ for \ this \ to \ work \\
\\
\text{otherwise we'll end up with a case where} \ 1/N= \epsilon
[/itex]
 

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  • #2
LCKurtz
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So use ##\epsilon/2## to begin with or let ##
N = \hbox{max}\{3,1/\epsilon\}+1##.
 
  • #3
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So use ##\epsilon/2## to begin with or let ##
N = \hbox{max}\{3,1/\epsilon\}+1##.
OK. So my question is correct, putting [itex]N = \hbox{max}\{3,1/\epsilon\}[/itex] was incorrect.
I found this argument on a website and wasn't sure.
Thanks.
 

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