Convergence of Sequence to e and around e

[mscudder3]

I was thinking of how ( 1 + (1/n) ) ^ n converges to e and I am aware of how if it is raised to some an, then it converges to e^a. If i recall if the form ( 1 + (a/n) ) ^ n converges to ae?

I was hoping someone could tell me how to deal with ( 1 + (1/n^2) ) ^ n?

Thanks!

 

[S_Happens]

Both of the first two sequences converge to e^a, not ae.

In the last example, the sequence converges to 1. I'll have to dig around for a minute to find something on it.

 

[mscudder3]

Both of the first two sequences converge to e^a, not ae.

In the last example, the sequence converges to 1. I'll have to dig around for a minute to find something on it.

I see. I just wanted some clarity on how the sequence acts when the degree of n in the (1/n) term is greater than the degree of the entire sequence. Is it fair to say that ( 1 + (a/n^2) ) ^ n converges to 1 as well, regardless of a?

(Thanks for looking into it!)

 

[S_Happens]

I'm sorry, I had made a mistake the first time around in my calculuation of the n^2, so I thought I didn't understand.

It's pretty simple to solve these limits. Take the natural log of the function to bring the exponent to the front, then force a fraction by putting the reciprocal (1/n) in the denominator. Use L'Hopital's rule and take individual derivatives. The limit that you get is the natural log of the limit, so you raise e to that power for the answer.

If you need to see it I'll have to play with Latex for a bit to get it right as I haven't used it much.

Edit- The (1 + 1/n^2)^n winds up as e^0, which is 1.

Also, the a constant doesn't matter in that example as well. If you work it out like I said, then you'll see that the constant is still in a fraction over the n, so it goes to 0. In the first examples, the n ends up cancelling out, which leaves the constant so that it does contribute to the limit.

 

[mscudder3]

Ya I'm not quite seeing how it works... it is probably due to my own error, let me elaborate.

taking the natural log and pulling down the n we get n * ln( 1 + (1/ n^2) ).
manipulating the n we get ln( 1 + (1/n^2) ) / (1/n)
applying L'Hopital's rule we get lim [(1/n^2) / ( 1 + (1/n^2) )] / (1/n^2)
cancel the 1/n^2... = lim 1 / (1 + (1/n^2) ) = 1
raising this to the power of e, we get e.

I must be overlooking something because in my work no matter the degree of the (1/n) in the interior, the result will always come out to 1.

Thanks in advance!

EDIT: I found my error, it was in the derivative of natural log lol, it has been far too long.

 

[S_Happens]

Yeah, the numerator after L'Hopital (derivative of the inside of the natural log function) is -2/n^3, so it doesn't cancel completely with what you have in the bottom.