# Quick sequence question

1. Nov 23, 2004

### kdinser

$$\frac{n}{2^{n+2}}$$

I know it's monotonic decreasing, a sub n < a sub n+1 and so has an upper bound of 1/8.

Can you then use L'Hopital's rule to determine that the sequence converges to 0, it's lower bound?

2. Nov 23, 2004

### Galileo

No. I don't think you'll ever have to use L'hospital to solve a sequence question. In this case you're not allowed to differentiate, since the function is only defined on the integers.

If a sequence is monotonic decreasing, then it is sufficient to show it has a LOWER bound for it to converge.

3. Nov 23, 2004

### HallsofIvy

Staff Emeritus
But $\frac{n}{2^{n+2}}$ has the same limit (as n-> infinity) as the continuous function $\frac{x}{2^{x+2}}$ (as x-> infinity). Certainly you can use L'Hopital to find the limit of the continuous function and then assert that as the limit of the sequence.

It is sufficient to show it has a lower bound (and 0 is an obvious lower bound) if you only want to prove that it does converge. That doesn't determine what it converges to.

4. Nov 23, 2004

### Galileo

Yes ofcourse, but you'll have to state that explicitly and not differentiate numerator and denominator w.r.t. n immediately.

In general, L'hospitals rule is a source for obtaining many fraudulent proofs. I'm just not a fan of using L'Hospital :yuck: