Sequences, convergance and limits.

In summary, the given series converges and the limit of the convergent sequences is 1. The limit can be calculated using the fact that m^(1/m) = e^(ln(m^(1/m))) and the L'Hospital rule. Taking the log of the expression and using the rules of logs, we can turn it into a quotient of infinity/infinity form. Then, we can use L'Hospital's rule to find the limit, which is 1.
  • #1
amplifierr
4
0

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.
 
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  • #2
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 until 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
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
amplifierr said:
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
amplifierr said:
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
(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
 

Related to Sequences, convergance and limits.

What is a sequence?

A sequence is a list of numbers that are arranged in a specific order. Each number in the sequence is called a term, and the position of a term in the sequence is called its index.

What is convergence?

Convergence refers to the behavior of a sequence as it approaches a certain value or limit. A sequence is said to converge if its terms become closer and closer to a specific value as the index increases.

How is the limit of a sequence determined?

The limit of a sequence is determined by observing the behavior of its terms as the index approaches infinity. If the terms of the sequence get closer and closer to a specific value as the index increases, then that value is the limit of the sequence.

What is a divergent sequence?

A divergent sequence is a sequence that does not have a limit. This means that its terms do not approach a specific value as the index increases.

How can we prove convergence of a sequence?

There are several methods to prove convergence of a sequence, such as the squeeze theorem, the monotone convergence theorem, and the Cauchy criterion. These methods involve analyzing the terms of the sequence and using mathematical techniques to show that they approach a specific value as the index increases.

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