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

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Discussion Overview

The discussion centers around the behavior of the sequence \( U_n \) as \( n \) approaches infinity, particularly in relation to the limits involving \( n U_n \) and their implications for inequalities. Participants explore the conditions under which \( U_n \) is greater than or less than \( \frac{A}{n} \), and the implications for the convergence or divergence of the series \( \sum U_n \).

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that \( \lim_{{n}\to{\infty}} {n}^{p}U_n \rightarrow A \) implies \( U_n < \frac{A}{{n}^{p}} \) for \( p > 1 \), but questions the reasoning behind \( \lim_{{n}\to{\infty}}n U_n = A \) leading to \( U_n > \frac{A}{n} \).
  • Another participant challenges the initial inequality, providing a counter-example with \( u_n = \frac{1}{n} \) and suggesting an alternative approach using \( \varepsilon \) to establish bounds for \( U_n \) for sufficiently large \( n \).
  • There is a discussion about using the comparison test to show divergence of \( \sum U_n \), with one participant asserting that \( U_n \le \frac{A}{n} \) contradicts the divergence conclusion, while another argues that \( U_n > \frac{A}{2n} \) leads to divergence.
  • One participant expresses confusion regarding the argument involving \( |nu_n - A| < \frac{1}{2}A \) and seeks clarification on its derivation.

Areas of Agreement / Disagreement

Participants do not reach consensus on the inequalities involving \( U_n \) and their implications for convergence or divergence of the series. Multiple competing views remain regarding the correct interpretation of the limits and the validity of the inequalities presented.

Contextual Notes

There are unresolved assumptions regarding the behavior of \( U_n \) and the specific conditions under which the inequalities hold. The discussion also reflects a dependence on the definitions and properties of the sequences involved.

ognik
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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 :-)
 
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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 said:
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?
 
ognik said:
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.
 
ThePerfectHacker said:
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?
 

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