Limit n[1 - exp(ia/n)]

1. Apr 8, 2008

e(ho0n3

[SOLVED] Limit n[1 - exp(ia/n)]

The problem statement, all variables and given/known data
What is the limit of n[1 - exp(ia/n)] as n goes to infinity (a is fixed real number).

The attempt at a solution
n[1 - exp(ia/n)] = n / [1 - exp(ia/n)]-1 and by using l'Hosptal's rule, I get that the limit diverges. Now how can I do this without relying on l'Hospital's rule?

2. Apr 8, 2008

tiny-tim

Hi e(ho0n3!

Are you sure you mean n[1 - exp(ia/n)], and not n[1 - exp(a/n)]?

That's n(1 - cos(a/n)) - i sin(a/n).

The real part oscillates between … ?, and the imaginary part oscillates between … ?

3. Apr 8, 2008

e(ho0n3

No.

That's correct.

I don't think it oscillates because a/n -> 0 as n -> infinity. 1 - exp(ia/n) -> 0 as n -> infinity. Or am I wrong?

4. Apr 9, 2008

tiny-tim

oops!

oh, sorry, I wasn't thinking straight.

Yes, so just expand exp(ia/n), or cos(a/n), as powers of a/n …

5. Apr 9, 2008

e(ho0n3

If I'm going to expand cos(a/n) as a power series, I might as well use l'Hospital's rule don't you think? And I don't see how this would help anyways.

6. Apr 9, 2008

tiny-tim

ah, but you only need the first couple of terms of the expansion (for either cos or exp)!

and where's the fraction that you'd apply l'Hôpital's rule (that's the official spelling!) to?

Alternatively, if you prefer, you could rewrite 1 - cos(a/n) in terms of sin(a/2n).

7. Apr 11, 2008

e(ho0n3

I know that but I don't understand how it is going to work. For example, the expansion of cos(z) about a/n is cos(a/n) + ... where ... are terms containing powers of z - a/n. Evaluating this at z = a/n gives me cos(a/n) again.

Please note that I don't want to use any method that relies on computing derivatives.

It's in my first post: n / [1 - exp(ia/n)]-1. (Wikipedia states he spelled his name with an s so I'm just following the man's preference.)

I'm not familiar with this identity. Would you care to explain?

8. Apr 11, 2008

Dick

I can't think of any way to do this that doesn't involve l'Hopital or derivatives (or series expansions, which are pretty much the same thing). But l'Hopital doesn't show that the series diverges. I will admit that the infinity/infinity form that you picked seems to just make the problem harder. Try (1-exp(ia/n))/(1/n).

9. Apr 11, 2008

tiny-tim

Hi e(ho0n3!

The expansion for cosθ is 1 - θ²/2 + θ^4/4! - θ^6/6! + …

(and the expansion for sinθ is θ - θ^3/3! + θ^5/5! - θ^7/7! + …)
Well, I never knew that!
Sure … you need to learn (though not necessarily for this question):
sin2θ = 2.sinθ.cosθ;
cos2θ = cos²θ - sin²θ;
1 + cos2θ = 2cos²θ;
1 - cos2θ = 2sin²θ;​
put θ = a/2n, and it's the last one.

(btw, if you did want to use l'Hôpital's rule, remember it's really only for 0/0, not for ∞/∞, so follow Dick's advice. )

10. Apr 11, 2008

Dick

l'Hopital can be used for either 0/0 or infinity/infinity forms. The only issue is that sometimes one form can lead to an easier solution than the other. Choosing the wrong form can sometimes complicate the problem rather than simplify it. If things don't seem to be going well, try and rearrange the indeterminant form to make life easier.

11. Apr 11, 2008

e(ho0n3

OK. I will satisfy myself with the derivation using l'Hospital's rule. Thank you tiny-tim and Dick.