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

In summary: We showed that $u_n > \tfrac{A}{2n}$ and $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, therefore $u_n$ diverges.
  • #1
ognik
643
2
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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  • #2
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 $N$.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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?
 

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

What is a limit comparison question?

A limit comparison question is a type of mathematical problem solving technique used to determine the behavior of a series or sequence. It involves comparing the given series to a known series with a known convergence or divergence behavior.

How do you solve a limit comparison question?

To solve a limit comparison question, you first need to identify the given series and the known series. Then, you take the limit of the quotient of the two series as the number of terms approaches infinity. If the limit is a finite positive number, the two series have the same behavior. If the limit is 0, the given series is smaller and will have the same behavior as the known series. If the limit is infinity, the given series is larger and will have the same behavior as the known series.

What are the conditions for using limit comparison?

The conditions for using limit comparison are that both series must have positive terms, and the known series must have a known convergence or divergence behavior. Additionally, the limit of the quotient of the two series must exist and be a finite positive number, 0, or infinity.

Can limit comparison be used to prove convergence or divergence?

Yes, limit comparison can be used to prove convergence or divergence of a given series. If the limit of the quotient of the two series is a finite positive number, then the given series has the same behavior as the known series and will converge or diverge based on the known series. If the limit is 0, the given series will have the same behavior as the known series and will converge. If the limit is infinity, the given series will have the same behavior as the known series and will diverge.

What are some common known series used in limit comparison?

Some common known series used in limit comparison include the harmonic series, geometric series, and p-series. These series have well-known convergence or divergence behaviors, making them useful for comparison to other series.

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