Does the sequence n/2^(n+2) converge to 0?

[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?

 

[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.

 

[HallsofIvy]

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.

 

[Galileo]

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.

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