Prove the sequence n/2^n converges to 0

1. Oct 22, 2009

naele

1. The problem statement, all variables and given/known data
Show that $$x_n=\frac{n}{2^n}$$ converges to 0

2. Relevant equations
Squeeze theorem?

3. The attempt at a solution
I've already proven that for $$n\geq 4, n^2 < 2^n$$ which means that $$\frac{1}{n} < \frac{n}{2^n}$$. My desired approach is to use the squeeze theorem, but I cannot think of another sequence greater than n/2^n that would converge to 0.

2. Oct 22, 2009

Staff: Mentor

Can you show that (n + 1)/2n + 1 < n/2n for all n past some initial value? (Proof by induction?) If you can show this, and then use the fact that the sequence is bounded below (by zero), you have sequence that is monotonically decreasing and bounded below by 0, hence converges to 0.

3. Oct 22, 2009

hamster143

You got the sign wrong. $$\frac{1}{n} > \frac{n}{2^n} > 0$$ for n>4. Next, apply the squeeze theorem.

that's not enough. For example, 1+1/n is monotonically decreasing and it's bounded below by 0, but it does not converge to 0.

4. Oct 22, 2009

Staff: Mentor

You got me. Mea culpa.

5. Oct 22, 2009

naele

Thanks hamster, my teacher warned me against this kind of mistake too. Live and learn I guess :)

6. Oct 22, 2009

HallsofIvy

Staff Emeritus
If $x_{n+1}/x_n$ has limit r< 1 then you can "compare" the sequence to $r^n$.