Limit demonstration of sequence

[joao_pimentel]

Hi guys

I know this limit goes to infinity

$$lim (3^n-n)$$
But how do I demonstrate it?

Actually I know also that this type of limits goes to infinity

$$lim \frac{a^n}{n^k},\forall a,k \in \mathbb{N},a>1$$
But I don't know how to prove it

May you kindly help me?

Many thanks

 

[DonAntonio]

Hi guys

I know this limit goes to infinity

$$lim (3^n-n)$$
But how do I demonstrate it?

Actually I know also that this type of limits goes to infinity

$$lim \frac{a^n}{n^k},\forall a,k \in \mathbb{N},a>1$$
But I don't know how to prove it

May you kindly help me?

Many thanks

One glamorous way to prove it:

1) this is a positive sequence;

2) the series $\displaystyle{\sum_{n=1}^\infty \frac{n^k}{a^n}}\,\,$ converges (D'alembert or Cauchy n-root test, say), thus $\,\,\displaystyle{\frac{n^k}{a^n}\to_{n\to\infty} 0}$ , so...

DonAntonio

 

[joao_pimentel]

Very interesting :) Really

Is there any other proof, more common and less glamorous (without going into series)?

 

[DonAntonio]

Very interesting :) Really

Is there any other proof, more common and less glamorous (without going into series)?

Another way: define $f(x):=\frac{a^x}{x^k}\,\,$, show this function has a min. at some point, after which the derivative is always

positive and thus the function's ascending, and since the function's always positive...

DonAntonio

 

[joao_pimentel]

But functions whose derivative is always positive after a certain minimum, don't mean they tend to infinity...

For example:

$$f(x)=-e^{-x}+1, x\geq 0$$
$$f(x)=-x, x<0$$

has a minimum at x=0 and after that is always ascending, though it tends to x=1

 

[DonAntonio]

But functions whose derivative is always positive after a certain minimum, don't mean they tend to infinity...


Neither did I say so nor even hinted at it. Please do read again what I wrote, in particular the "always positive" thingy.

DonAntonio ***


For example:

$$f(x)=-e^{-x}+1, x\geq 0$$
$$f(x)=-x, x<0$$

has a minimum at x=0 and after that is always ascending, though it tends to x=1

...

 

[joao_pimentel]

Maybe with the second derivative which gives the concavity...

 

[joao_pimentel]

Thank you DonAntonio

I really apologise but I can't see the big picture...

Facts:

1. The function has a minimum at certain point
2. After that minimum the derivative is always positive
3. The function is always positive, i.e. f(x)>0, for all x in R

How do I conclude that it tends to infinity?

Sorry to bother...

Thank you very much

 

[DonAntonio]

Thank you DonAntonio

I really apologise but I can't see the big picture...

Facts:

1. The function has a minimum at certain point
2. After that minimum the derivative is always positive
3. The function is always positive, i.e. f(x)>0, for all x in R

How do I conclude that it tends to infinity?

Sorry to bother...

Thank you very much

Well, I intended for you to complete the picture: since the function I used has no oblicuous and/or horizontal asymptotes, and since it is

always positive, if after some definite point is derivative is positive then the function must an ascending after that point, so

it MUST tend to $$\infty$$

DonAntonio

 

[joao_pimentel]

Thank you very much DonAntonio

I confess I was not reaching the part of the asymptotes...

Thank you so very much indeed

Greetings from Lisbon

João

 

[DonAntonio]

Thank you very much DonAntonio

I confess I was not reaching the part of the asymptotes...

Thank you so very much indeed

Greetings from Lisbon

João

Foi un prazer para mim.

DonAntonio

 

[joao_pimentel]

Muito obrigado mesmo pela atenção

Melhores cumprimentos

 

[Office_Shredder]

The way I like to think about these kinds of ratios (aⁿ/n^k) is by using this:
$$\lim_{n\to \infty} \left( \frac{n+1}{n} \right)^k = 1$$
So when n is a large enough number, aⁿ/n^k might be very small, we don't know. But each time I increase n by one after this point, the numerator increases by a factor of a, and the denominator increases by a factor which is very close to 1 (say, less than a^1/2). From that point on the value of aⁿ/n^k grows by a factor of at least a^1/2 each time you increase n by one

 

[joao_pimentel]

Very interesting thought, thank you very much...