# Sequence infinity proof

1. Jan 8, 2009

1. The problem statement, all variables and given/known data

Prove that the following sequence (a(n)) has the property that a(n) tends to infinity as n tends to infinity.

2. Relevant equations
a(n)=[n+7]/[2+sin(n)]

3. The attempt at a solution

i tried l'hopitals rule, so i got 1/cos(n).....which wouldnt work.
so im not sure how to do this question.

2. Jan 8, 2009

### Staff: Mentor

Re: proof

L'Hopital's Rule doesn't apply here. The quotient has to be of the form 0/0 or infinity/infinity for L'H's Rule to be used.

Instead, use the fact that 1 <= 2 + sin(n) <= 3, for all n to get lower and upper limits on the value of (n + 7)/(2 + sin(n)), and then show that the lower and upper limit expressions both have the same limit as n gets large, thereby trapping the expression in the middle.

3. Jan 8, 2009

### Dick

Re: proof

Since n+7 is unbounded, to apply l'Hopital to this you would need that the denominator is also unbounded (so you have an infinity/infinity form). It's not. The denominator is bounded and positive. That should tell you the limit right there.