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Sequences, convergance and limits.

  • Thread starter amplifierr
  • Start date
  • #1

Homework Statement


An = (n+3)^1/(n+3)
Converge or diverges? Find the limit of the convergent sequences.

Homework Equations





The Attempt at a Solution



I have the solution but I don't understand it. I'm looking to get it.

It shows me taking the limit as x->infinity of x^1/x and determine that it is 1. (Thus the series converges). I am not sure at all how to find the limits though, I'm lost in my calculus class. Any guidance would be great.
 

Answers and Replies

  • #2
938
9
Use the fact that [tex] m^{1/m} = e^{\ln (m^{1/m})} [/tex] and then for example the L'Hospital rule to calculate the limit for a large m.
 
  • #3
So my f(x) would be my original

and the g(x) would be my e^lnn(n+3)^(1/n+3)

I put f'(x) over g'(x) and keep going untill it's not 0/0, whatever I do get beside 0/0 is my limit?

I'm honestly stuck trying to differentiate my g(x) ;X Is this the only way to go about this?
 
  • #4
Dick
Science Advisor
Homework Helper
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618
Take the log of your expression. Then you should get something of the form f(n)/g(n). You won't get that till you take the log.
 
  • #5
log of n+3?

I don't see how that gets me anywhere?
 
  • #6
Dick
Science Advisor
Homework Helper
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618
log of n+3?

I don't see how that gets me anywhere?
No, log((n+3)^(1/(n+3))).
 
  • #7
you want me to derive that then? Sorry I'm lost at the moment.
 
  • #8
Dick
Science Advisor
Homework Helper
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you want me to derive that then? Sorry I'm lost at the moment.
I want you to use the rules of logs and turn that into a quotient of infinity/infinity form, then use l'Hopital.
 
  • #9
24
0
(n+3)^[1/(n+3)] is just the same as saying x^(1/x).

x^(1/x) = e^(1/x lnx)

the limit as x -> infinity, lim x^(1/x) = lim e^(1/x lnx) = e^lim (lnx / x)

lim (lnx / x) = lim (1/x) {Le Hopital's rule}
= 0

Thus, e^0 = 1

and the series is convergent
 

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