# Sequences, convergance and limits.

1. Nov 24, 2009

### amplifierr

1. The problem statement, all variables and given/known data
An = (n+3)^1/(n+3)
Converge or diverges? Find the limit of the convergent sequences.

2. Relevant equations

3. 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.

2. Nov 24, 2009

### clamtrox

Use the fact that $$m^{1/m} = e^{\ln (m^{1/m})}$$ and then for example the L'Hospital rule to calculate the limit for a large m.

3. Nov 25, 2009

### amplifierr

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. Nov 25, 2009

### Dick

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. Nov 25, 2009

### amplifierr

log of n+3?

I don't see how that gets me anywhere?

6. Nov 25, 2009

### Dick

No, log((n+3)^(1/(n+3))).

7. Nov 25, 2009

### amplifierr

you want me to derive that then? Sorry I'm lost at the moment.

8. Nov 25, 2009

### Dick

I want you to use the rules of logs and turn that into a quotient of infinity/infinity form, then use l'Hopital.

9. Nov 27, 2009

### andylu224

(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