Does the Sequence sqrt[n]{3^n + 5^n} Converge or Diverge?

[grothem]

Homework Statement

Determine if the sequence is convergent or divergent
{\sqrt[n]{3^n+5^n}}

Homework Equations

The Attempt at a Solution

I know I need to take the limit to find if it converges or diverges. But I'm not really sure what I need to do to it to take the limit.

 

[Dick]

Take the log of the sequence. Write 3^n+5^n=5^n(1+(3/5)^n). Does that help?

 

[th3plan]

no need to that even, use the root test, and from there you can tell

 

[grothem]

I thought you could only apply the root test to a series, not a sequence, or does it not matter?

 

[Dick]

I thought you could only apply the root test to a series, not a sequence, or does it not matter?

Right. The root test is not helpful for sequences.

 

[grothem]

so after taking the log I came up with 5(1+(3/5)) = 8. Which I don't think is the right answer, did I do something wrong?

 

[Dick]

so after taking the log I came up with 5(1+(3/5)) = 8. Which I don't think is the right answer, did I do something wrong?

Yes, quite a bit. Review your rules of logs. Try again. log([5^n*(1+3^n/5^2)]^(1/n)). What's your first step?

 

[grothem]

oh ok...

so I bring the 1/n in front of the ln(5^n*(3/5)^n)

now can I just take the limit from there?

 

[Dick]

Careful, you dropped a very important '1'. Now use log(a*b)=log(a)+log(b). Then the power rule again.

 

[grothem]

I came up with Limit as n tends to infinity of (ln(3)*(3/8)^n + ln(5)*(5/8)^n)
and with those being geometric sequences with r < 1, the sequence is convergent.

 

[Dick]

I came up with Limit as n tends to infinity of (ln(3)*(3/8)^n + ln(5)*(5/8)^n)
and with those being geometric sequences with r < 1, the sequence is convergent.

That's a long way from being correct. How did you get that?

 

[grothem]

I took the natural log and I got:

Lim 1/n * ln(3^n+5^n)

Then applying L'hospitals:

(3^n*ln(3)+5^n*ln(5))/(3^n+5^n)

then I simplified from there.

 

[Dick]

I took the natural log and I got:

Lim 1/n * ln(3^n+5^n)

Then applying L'hospitals:

(3^n*ln(3)+5^n*ln(5))/(3^n+5^n)

then I simplified from there.

If you want to go that way then fine. Then the 'simplification' went wrong. Try dividing numerator and denominator by 5^n. What's the limit of 3^n/5^n?

 

[grothem]

If you want to go that way then fine. Then the 'simplification' went wrong. Try dividing numerator and denominator by 5^n. What's the limit of 3^n/5^n?

ok. The limit of 3^n/5^n = 0

So after dividing through by 5^n, I'm left with ln(5). Which would be a divergent series

 

[Dick]

ok. The limit of 3^n/5^n = 0

So after dividing through by 5^n, I'm left with ln(5). Which would be a divergent series

It's a divergent series, but it's a convergent sequence. It converges to ln(5). But now remember you took the log of the original sequence.

 

[grothem]

It's a divergent series, but it's a convergent sequence. It converges to ln(5). But now remember you took the log of the original sequence.

Oh ok. So raising ln(5) to power e gives me 5. So sequence converges to 5.
Thanks a lot for the help!

 

[Dick]

You're welcome. I think it would be a good exercise to try and do this without l'Hopital. You don't really need it, just use the rules of logs.

 

[jlu]

how can i proof that ([sin0.4n][/npi])2 converges and ([sin0.4n][/npi]) diverges? n is between -infinity and +infinity

 

[dirk_mec1]

What is the sequence? It isn't clear in the way you've written down.

 

[Dick]

how can i proof that ([sin0.4n][/npi])2 converges and ([sin0.4n][/npi]) diverges? n is between -infinity and +infinity

Open a new thread for a new question. Don't tag it onto an old thread. It won't get the attention it deserves. And try to post the question more legibly on the new thread. What is ([sin0.4n][/npi])2? Do you mean (sin(theta*4n)/(n*pi))^2??