Using arithmetic of limits

[CStudent]

Hey.

I have some exercise in CALCULUS, I haven't succeeded to solve:

For all x>0 and Natural n we get: $(1+x)^n>\frac{n(n-1)}{2}x^2$

Prove that if $1<q\in \mathbb{R}$ is some constant value then: $\displaystyle{\lim_{n \to \infty}\frac{n}{q^n}}=0$

I actually tried to start by the inequality as I know the left side is from the identity of Bernoullis inequality but it led me nowhere...

Thanks.

 

[Plato]

Prove that if $1<q\in \mathbb{R}$ is some constant value then: $\large\color{blue}{\displaystyle{\lim_{n \to \infty}\frac{n}{q^n}}=0}$
I actually tried to start by the inequality as I know the left side is from the identity of Bernoullis inequality but it led me nowhere...

I an confused is that what you want to prove?

 

[CStudent]

I an confused is that what you want to prove?

Yes

 

[Plato]

Yes

So we are put at a disadvantage. We have no idea what theorems and/or axioms/definitions you have to work with.
This is a fairly standard result but it does come in the sequence of well developed set of theorems.
Here is an example.
If $q>1$ then $q=(1+x)$ so
\(\displaystyle \begin{align*}q^n=(1+x)^n&\ge\dfrac{n(n-1)x^2}{2} \\\dfrac{1}{q^n}&\le\dfrac{2}{n(n-1)x^2}\\\dfrac{n}{q^n}&\le\dfrac{2}{(n-1)x^2} \end{align*}\)

 

[CStudent]

So we are put at a disadvantage. We have no idea what theorems and/or axioms/definitions you have to work with.
This is a fairly standard result but it does come in the sequence of well developed set of theorems.
Here is an example.
If $q>1$ then $q=(1+x)$ so
\(\displaystyle \begin{align*}q^n=(1+x)^n&\ge\dfrac{n(n-1)x^2}{2} \\\dfrac{1}{q^n}&\le\dfrac{2}{n(n-1)x^2}\\\dfrac{n}{q^n}&\le\dfrac{2}{(n-1)x^2} \end{align*}\)

Hey thank you.
We have reached to the Cantor's lemma.

A few questions on your solution:
How are we sure that n is not equal 1?

How do we know from this inequality now that the limit of the sequence n/q^n is 0? According to what?
We know that 2 is a constant sequence so its limit is 2, we know that x^2 is also constant so its limit is x^2, and (n-1) tends to infinity (Do I have to prove it somehow?) - So we have 2 at the nominator and infinity at the denominator - which theorem tells me that yields 0?

And we need to pose another 0 left to n/q^n to use the sandwich, but how do we do that?

Thank you!

 

[Plato]

Hey thank you.
We have reached to the Cantor's lemma.
A few questions on your solution:
How are we sure that n is not equal 1?
How do we know from this inequality now that the limit of the sequence n/q^n is 0? According to what?
We know that 2 is a constant sequence so its limit is 2, we know that x^2 is also constant so its limit is x^2, and (n-1) tends to infinity (Do I have to prove it somehow?) - So we have 2 at the nominator and infinity at the denominator - which theorem tells me that yields 0?

And we need to pose another 0 left to n/q^n to use the sandwich, but how do we do that?

How are we sure that n is not equal 1? $n\to \infty$ so what does it matter?
It seem that you have answered your own questions.
$0< \dfrac{n}{q^n}\le\dfrac{2}{(n-1)x^2}\to 0$

 

[CStudent]

How are we sure that n is not equal 1? $n\to \infty$ so what does it matter?
It seem that you have answered your own questions.
$0< \dfrac{n}{q^n}\le\dfrac{2}{(n-1)x^2}\to 0$

Why we can pose another 0 left to n/q^n? What enables this step?
And how do I explain that the very right fraction is also tending to 0?

Thank you!

 

[Plato]

Why we can pose another 0 left to n/q^n? What enables this step?

If you do no understand that $q>0~\&~n\to\infty$ implies that $\dfrac{n}{p^n}>0$ you are wasting time with this.

And how do I explain that the very right fraction is also tending to 0?

This in most elementary of limits: if $A~\&~B$ are non-zero constants then \(\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{A}{{B \cdot n}} = 0\).
Again if that is not clear to you the don't waste your time.

 

[CStudent]

If you do no understand that $q>0~\&~n\to\infty$ implies that $\dfrac{n}{p^n}> you are wasting time with this.


This in most elementary of limits: if $A~\&~B$ are non-zero constants then \(\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{A}{{B \cdot n}} = 0\).
Again if that is not clear to you the don't waste your time.

Implies that what?
Okay thanks!