Convergence of a sequence

  • Thread starter grothem
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  • #1
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Homework Statement


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


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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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Take the log of the sequence. Write 3^n+5^n=5^n(1+(3/5)^n). Does that help?
 
  • #3
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no need to that even, use the root test, and from there you can tell
 
  • #4
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I thought you could only apply the root test to a series, not a sequence, or does it not matter?
 
  • #5
Dick
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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.
 
  • #6
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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?
 
  • #7
Dick
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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?
 
  • #8
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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?
 
  • #9
Dick
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Careful, you dropped a very important '1'. Now use log(a*b)=log(a)+log(b). Then the power rule again.
 
  • #10
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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.
 
  • #11
Dick
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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?
 
  • #12
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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.
 
  • #13
Dick
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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?
 
  • #14
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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
 
  • #15
Dick
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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.
 
  • #16
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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 alot for the help!
 
  • #17
Dick
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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.
 
  • #18
jlu
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how can i proof that ([sin0.4n][/npi])2 converges and ([sin0.4n][/npi]) diverges? n is between -infinity and +infinity
 
  • #19
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What is the sequence? It isn't clear in the way you've written down.
 
  • #20
Dick
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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??
 

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