Lim as n approaches infinity for n/2^n

[Mosaness]

1. Determine whether the sequence converges or diverges. If it converges, find its limit.

{\frac{n}{2ⁿ}}^+∞_{n = 1}

I know that we have to use L'Hopital's rule for this, because as n increases, both the numerator and the denomator approach infinity. However, doing that gives me \frac{1}{n*2^n-1}, which complicates things further.

I have a calc II quiz on Monday, so any help would be appreciated.

 

[Mark44]

1. Determine whether the sequence converges or diverges. If it converges, find its limit.

{\frac{n}{2ⁿ}}^+∞_{n = 1}

I know that we have to use L'Hopital's rule for this, because as n increases, both the numerator and the denomator approach infinity. However, doing that gives me \frac{1}{n*2^n-1}, which complicates things further.

I have a calc II quiz on Monday, so any help would be appreciated.

You are not differentiating 2ⁿ correctly. You are using the power rule, but 2ⁿ is an exponential function, not a power function. It's true that d/dx(x³) = 3x², but it is NOT true that d/dx(2^x) = x*2^{x - 1}, which is what you did.

Rewrite 2ⁿ as e^n*ln(2)

 

[Mosaness]

Okay. I know that \frac{dx}{dy} of e^x is e^x.

So for 2ⁿ, it can be rewritten as e^n*ln2.

Let u = n*ln 2, so e^n*ln2 becomes e^u.

Differentiating this becomes e^u.

And differentiating u should...be done using the product rule correct?

Which should give (1)(ln 2) because differentiating ln 2 becomes 0.

This should give us a derivative, overall, of e^{ln 2}.

Am I correct? I apologize in advance for any mistakes I might have made.

 

[Mark44]

Okay. I know that \frac{dx}{dy} of e^x is e^x.

Not quite. You don't take dx/dy of something (or dy/dx of something either).
If y = e^x, then dy/dx = e^x.

So for 2ⁿ, it can be rewritten as e^n*ln2.

Let u = n*ln 2, so e^n*ln2 becomes e^u.

Differentiating this becomes e^u.

You need to be more specific. Differentiating with respect to what?

d/du(e^u) = e^u, but to get d/dn(e^u) you need to use the chain rule.

So d/dn(e^u) = e^u * du/dn.

And differentiating u should...be done using the product rule correct?

You could, but that's more work than necessary. The constant multiple rule would be the right one. To differentiate 3x, the product rule would work, but it's simpler to do this: d/dx(3x) = 3*d/dx(x) = 3 * 1 = 3.

Which should give (1)(ln 2) because differentiating ln 2 becomes 0.

This should give us a derivative, overall, of e^{ln 2}.

No.

Am I correct? I apologize in advance for any mistakes I might have made.

 

[Mosaness]

What is the correct answer then?

 

[Mosaness]

y = e^u
where u = n*ln2

dy/du = e^u
du/dx = ln 2?

Therefore
dy/dx = ln 2*e^{n*ln 2}?

 

[klondike]

Then write e^{n*ln 2} back to the simpler form.

y = e^u
where u = n*ln2

dy/du = e^u
du/dx = ln 2?

Therefore
dy/dx = ln 2*e^{n*ln 2}?

 

[Mosaness]

Which becomes ln 2 * 2ⁿ.

When you take the limit of n/2^n, using L'Hopital's rule, you get 1/ln2*2ⁿ.

This converges to 0, because as n gets bigger, the function overall will get closer and closer to 0?

COrrect?

 

[klondike]

That's correct. You can further squeeze yourself by using ε-N to prove it:)

Which becomes ln 2 * 2ⁿ.

When you take the limit of n/2^n, using L'Hopital's rule, you get 1/ln2*2ⁿ.

This converges to 0, because as n gets bigger, the function overall will get closer and closer to 0?

COrrect?

 

[Mosaness]

How would we do that? We were explained the theorem in class, I didn't understand it fully however.

 

[Mark44]

The limit of your sequence is 0. The work you did is probably sufficient unless you were asked to use an epsilon-type argument to prove that the limit is zero.

I'm assuming that you weren't asked to do it this way.

 

[Mosaness]

Well, it was a part of a homework problem, but should he ask how to do it using that argument, how would that be done?

 

[Mark44]

Well, it was a part of a homework problem, but should he ask how to do it using that argument, how would that be done?

Why don't you wait until he actually asks? I'm sure your instructor will show you some examples.

 

[klondike]

Well, I looked at it further and found that the ε-N proof in this case would not be straightforward.

You want \frac{n}{2^n}<\epsilon, it's not obvious how to solve the inequality. We want to reasonably magnifiy the LHS a bit to a nicer function. This may be done by invoking mean value theorem.

\frac{2^n-1}{n}=2^{\zeta}ln2 \quad ; \quad 0<\zeta<n. It follows that

\frac{n}{2^n}<\frac{1}{2^{\zeta}ln2}
Now it looks much better.

Well, it was a part of a homework problem, but should he ask how to do it using that argument, how would that be done?

 

[klondike]

I thought about it a bit more, to do ε-N proof, I would have to accept the fact that 2^n>n^2 for n>4. I don't know how to algebraically prove it though:shy:

With that given, the epsilon-N proof can be done by:

We want \left|\frac{n}{2^n}\right|<\epsilon
Notice that
\frac{n}{2^n}<\frac{n}{n^2}=\frac{1}{n}<\epsilon \quad for \quad n>4

For any ε>0, there exists integer N=max(4,[1/ε]) such that if n>N, then \left|\frac{n}{2^n}\right|<\epsilon
QED

 

[206PiruBlood]

I thought about it a bit more, to do ε-N proof, I would have to accept the fact that 2^n>n^2 for n>4. I don't know how to algebraically prove it though:shy:

Have you tried induction? You can also do 2^n≥n^2 to include n=4.