Convergence of sequence!

[evinda]

Hello!
I want to check if the sequence \(\displaystyle a_{n}=\frac{n}{a^{n}} , a>1 \) converge.
How can I do this? I got stuck right now...

 

[pka]

I want to check if the sequence \(\displaystyle a_{n}=\dfrac{n}{a^{n}} , a>1 \) converge.
How can I do this? I got stuck right now...

This question is quite clearly designed for the use of the root test.
\(\displaystyle \displaystyle\sqrt[n]{{{a_n}}} = \dfrac{{\sqrt[n]{n}}}{a}\)

 

[evinda]

This question is quite clearly designed for the use of the root test.
\(\displaystyle \displaystyle\sqrt[n]{{{a_n}}} = \dfrac{{\sqrt[n]{n}}}{a}\)

So,the sequence diverges??
I tried to check if it converge,by finding the monotonicity and I showed that it is decreasing and then I wrote that it is bounded above..Am I wrong???

 

[evinda]

This question is quite clearly designed for the use of the root test.
\(\displaystyle \displaystyle\sqrt[n]{{{a_n}}} = \dfrac{{\sqrt[n]{n}}}{a}\)

Can we apply the root test also at sequences?I thought that it would be just for series...

 

[pka]

Can we apply the root test also at sequences?I thought that it would be just for series...

It is for series, so that tells us \(\displaystyle \sum {{a_n}} \) converges.

So what does that tell you about \(\displaystyle \left( {{a_n}} \right) \to ~?\)

 

[evinda]

It is for series, so that tells us \(\displaystyle \sum {{a_n}} \) converges.

So what does that tell you about \(\displaystyle \left( {{a_n}} \right) \to ~?\)

Maybe that \(\displaystyle a_{n}->0 \) ??Is the way I did it wrong??

 

[pka]

Maybe that \(\displaystyle a_{n}->0 \) CORRECT!

Is the way I did it wrong??

How are we supposed to answer that?
You did not post any work.

 

[evinda]

How are we supposed to answer that?
You did not post any work.

I wrote at a previous post that I tried to check if it converge,by finding the monotonicity and I showed that it is decreasing and then I wrote that it is bounded above..Am I wrong??? :???:

 

[pka]

I wrote at a previous post that I tried to check if it converge,by finding the monotonicity and I showed that it is decreasing and then I wrote that it is bounded above..Am I wrong??? :???:

If it is decreasing then being bounded above has nothing to do with it.
It is bounded below by 0.

 

[evinda]

If it is decreasing then being bounded above has nothing to do with it.
It is bounded below by 0.

Oh!!!I am sorry.. :roll: I meant that it is bounded below..So,can I say that the sequence is decreasing and bounded below,so it converges?

 

[daon2]

Just a different approach: Can you find \(\displaystyle \displaystyle \lim_{x\to\infty} \dfrac{x}{a^x}\). Hint: L'Hopitals.

If you show that, then \(\displaystyle a_n = f(n)\) and \(\displaystyle f(n)\to L\) implies \(\displaystyle a_n\to L\).

The monotone convergence theorem tells you that it does converge but not what it converges to. Your use of the theorem is not right though. Decreasing and bounded below imply convergence, not decreasing and bounded above.

 

[evinda]

Just a different approach: Can you find \(\displaystyle \displaystyle \lim_{x\to\infty} \dfrac{x}{a^x}\). Hint: L'Hopitals.

If you show that, then \(\displaystyle a_n = f(n)\) and \(\displaystyle f(n)\to L\) implies \(\displaystyle a_n\to L\).

The monotone convergence theorem tells you that it does converge but not what it converges to. Your use of the theorem is not right though. Decreasing and bounded below imply convergence, not decreasing and bounded above.

\(\displaystyle \displaystyle \lim_{x\to\infty} \dfrac{x}{a^x}=0\) .Right?
So,if I firstly want to show that it converges,can I say that it does because of the fact that it is decreasing and bounded below?

 

[evinda]

I don't understand you sometimes. You post very difficult problems on here so you must be at a pretty advanced level. But then you ask if a monotonic decreasing sequence bounded from below converges?

I know that this theorem is valid,but I wanted to know if I can apply this at the specific exercise..

 

[daon2]

I know that this theorem is valid,but I wanted to know if I can apply this at the specific exercise..

yes the sequence is decreasing but only for n> 3. It is a positive sequence and so bounded below.