# Limit of a sequence

1. Feb 1, 2008

### fk378

1. The problem statement, all variables and given/known data
Determine whether the sequence converges or diverges. If it converges, what does it converge to?

(cos[n])^2 / (2^n)

2. Relevant equations
L'Hospital's Rule
or
Squeeze Theorem

3. The attempt at a solution
As n-->infinity, this function approaches infinity/infinity. Applying L'Hospital's rule gives -(sin[n])^2 / 2 which gives infinity/2=infinity.

However, if I use the squeeze theorem to say that 0<(cos[n])^2<1 and then divide everything by 2^n, then I can say that this function approaches 0.

Which is the correct method?

2. Feb 1, 2008

### Vid

The derivative of $$2^{n} = ln(2)*2^{n}$$, and the derivative of $$cos^{2}(n) = -2cos(n)sin(n).$$

Last edited: Feb 1, 2008
3. Feb 1, 2008

### Dick

i) The sequence isn't infinity/infinity. You just said 0<=cos(n)^2<=1. How can it be infinity/infinity?? So you can't use l'Hopital. 2) The derivative of 2^n is NOT 2. The squeeze argument is correct.

4. Feb 1, 2008

### HallsofIvy

Staff Emeritus
I would also point out that $\lim_{n\rightarrow \infty} sin(n)$ is NOT infinity. It just doesn't exist.