Sunday 19 July 2015

convergence divergence - How to prove an increasing sequence that converges is bounded above by its limit



I am trying to prove that an increasing sequence that converges to $ L$ is bounded above by its limit.
By using $a_n \le a_{n+1}$ and the definition of limit of a sequence, I can prove that for $\epsilon > 0$ , $ a_n \lt {L + \epsilon} $ for all $a_n$.
But is there a way to proceed to $ a_n \le L $ ? because I can't think of a case in which the former is true but the latter isn't.


Answer



HINT




You can easily show that if for some n $a_n>L$ then by definition of limit $a_n$ must decrease which is impossible.



You only need to formalize this idea by setting “assume exists n such that ...then by definition of limit...contradiction”.



Notably




  • suppose $\exists n_1$ such that $a_{n_1}>L$ with $d=a_{n_1}-L>0$

  • set $\epsilon=d$ by definition of limit must exists $n_2>n_1$ such that $|a_{n_2} -L|<\epsilon \implies a_{n_2}


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...