- #1
converting1
- 65
- 0
Use only the definition of convergence to prove ## \dfrac{cos(n)}{n} \rightarrow 0 ##
my proof:
given ## \epsilon > 0 ## ##\exists N \in \mathbb{R} ## s.t. n>N => ## |\dfrac{cos(n)}{n}| < \epsilon ##
## |cos(n)| \leq 1 ## therefore ## \dfrac{|cos(n)|}{|n|} \leq \dfrac{1}{|n|} < \epsilon ## so ## |n| > \dfrac{1}{\epsilon} ##
so choose big N to be 1/e and we get ## |\dfrac{cos(n)}{n}| < \epsilon ##
is this proof ok?
I am also confused about this
from here: ## \dfrac{|cos(n)|}{|n|} < \epsilon ## we can do ## -\epsilon |n| < cos(n) < \epsilon |n| ## and if we consider ## cos(n) < \epsilon |n| ## we get ## \epsilon |n| > -1 ## and therefore ## n > \dfrac{-1}{\epsilon} ## and ## n < \dfrac{1}{\epsilon} ## why do I get two different results contradicting with my first proof, I'm assuming I've gone wrong so if someone could correct me where, please.
my proof:
given ## \epsilon > 0 ## ##\exists N \in \mathbb{R} ## s.t. n>N => ## |\dfrac{cos(n)}{n}| < \epsilon ##
## |cos(n)| \leq 1 ## therefore ## \dfrac{|cos(n)|}{|n|} \leq \dfrac{1}{|n|} < \epsilon ## so ## |n| > \dfrac{1}{\epsilon} ##
so choose big N to be 1/e and we get ## |\dfrac{cos(n)}{n}| < \epsilon ##
is this proof ok?
I am also confused about this
from here: ## \dfrac{|cos(n)|}{|n|} < \epsilon ## we can do ## -\epsilon |n| < cos(n) < \epsilon |n| ## and if we consider ## cos(n) < \epsilon |n| ## we get ## \epsilon |n| > -1 ## and therefore ## n > \dfrac{-1}{\epsilon} ## and ## n < \dfrac{1}{\epsilon} ## why do I get two different results contradicting with my first proof, I'm assuming I've gone wrong so if someone could correct me where, please.