Hi, can anyone give me a hand with this question, i honestly dont know what to do!!
Prove that lim (n tends to infinity) 1/an = 0 when an is a sequence with an > 0 for n being natural, and lim (n tends to infinity) an = infinity.
Hi, can anyone give me a hand with this question, i honestly dont know what to do!!
Prove that lim (n tends to infinity) 1/an = 0 when an is a sequence with an > 0 for n being natural, and lim (n tends to infinity) an = infinity.
$\displaystyle \forall \,\epsilon >0,\,\exists \,k\in \mathbb{N}\mid n\ge k\implies \left| \frac1n\right|<\epsilon .$
But this is quite straightforward, so we get that $\displaystyle \frac1n<\epsilon$ which means that $\displaystyle n>\frac1\epsilon,$ so for any $\displaystyle k\in\mathbb N$ being greater than $\displaystyle \frac1\epsilon,$ the above condition is fulfilled. For example, it's enough to take $\displaystyle k=\left\lfloor \frac{1}{\epsilon} \right\rfloor +1.$