Why is $U_n$ greater than $\frac{A}{n}$ for large enough values of $n$?

[ognik]

Hi, my book says that $\lim_{{n}\to{\infty}} {n}^{p}U_n \rightarrow A \lt \infty, p \gt 1 $ means that $U_n \lt \frac{A}{{n}^{p}} $, which I can see

But apparently $ \lim_{{n}\to{\infty}}n U_n = A \gt 0 $ means that $ U_n \gt \frac{A}{n} $ I know this is going to sound like a stupid question, but please walk me through why this is? My head is stuck in a loop and maybe why I think I understand the 1st one is wrong as well... Thanks for patience :-)

 

[ThePerfectHacker]

Your inequality is not correct as stated. Take $u_n = \frac{1}{n}$ as counter-example. See it does not work.

Instead here is an alternative one.

Let $\varepsilon = \frac{A}{2} > 0$. There exists an $N>0$ large enough so that if $n>N$ then $|nu_n - A| < \frac{1}{2}A$. In particular, $\frac{1}{2}A< nu_n < \frac{3}{2}A$ and so we see that $u_n > \frac{A}{2n}$. Not for all $n$, but those those large enough and beyond $N$.

 

[ognik]

But apparently $ \lim_{{n}\to{\infty}}n U_n = A \gt 0 $ means that $ U_n \gt \frac{A}{n} $

So $nU_n \le A, \therefore U_n \le \frac{A}{n}$ , but then how do I show - using the comparison test - that $\sum_{}^{} U_n$ diverges? Unless there is a typo in the book 'cos it seems to me to converge?

 

[ThePerfectHacker]

So $nU_n \le A, \therefore U_n \le \frac{A}{n}$ , but then how do I show - using the comparison test - that $\sum_{}^{} U_n$ diverges? Unless there is a typo in the book 'cos it seems to me to converge?

We showed that $u_n > \tfrac{A}{2n}$ and $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, therefore $u_n$ diverges.

 

[ognik]

Your inequality is not correct as stated. Take $u_n = \frac{1}{n}$ as counter-example. See it does not work.

Instead here is an alternative one.

Let $\varepsilon = \frac{A}{2} > 0$. There exists an $N>0$ large enough so that if $n>N$ then $|nu_n - A| < \frac{1}{2}A$. In particular, $\frac{1}{2}A< nu_n < \frac{3}{2}A$ and so we see that $u_n > \frac{A}{2n}$. Not for all $n$, but those those large enough and beyond $N$.

Sorry, I confess I couldn't follow your argument.
I understand what some N, n > N means, but I don't see where the step $|nu_n - A| < \frac{1}{2}A$ comes from?