Prove the sequence n/2^n converges to 0

[naele]

Homework Statement

Show that x_n=\frac{n}{2^n} converges to 0

Homework Equations

Squeeze theorem?

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.

 

[Mark44]

Can you show that (n + 1)/2^{n + 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.

 

[hamster143]

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

you have sequence that is monotonically decreasing and bounded below by 0, hence converges to 0.

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.

 

[Mark44]

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.

You got me. Mea culpa.

 

[naele]

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

 

[HallsofIvy]

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