# (n )/(2 ^n ) sequence limit problem

1. Oct 3, 2004

Hello all

If I have a sequence: (n )/(2 ^n )

and I have already proved the lim (n)/ (2^n) = 0
n->00

then how would I find a number N such that (n)/ (2^n) < 1/ 10 ( or any other epsilon)? In other words how would I solve this equation?

Any help would be greatly appreciated

Thanks

2. Oct 3, 2004

### e(ho0n3

You can't really solve this kind of equation (at least to my knowledge). Looks like you're stuck using trial and error.

3. Oct 5, 2004

But that would take a long time (its in Courants book)

4. Oct 5, 2004

### arildno

Prove by induction that for n>=4, $$2^{\frac{n}{2}}\geq{n}$$
1. n=4:
$$2^{\frac{4}{2}}=4\geq4$$
2. Let it be true for n=k>=4.
Then,
$$2^{\frac{k+1}{2}}=\sqrt{2}2^{\frac{k}{2}}\geq(1+\frac{1}{4})k\geq(k+1)$$
QED

5. Oct 5, 2004

Yea but I am trying to solve for n not prove that its true

6. Oct 5, 2004

### arildno

That was not the point!
Were you unable to see the trivial consequence:
$$\frac{n}{2^{n}}\leq\frac{2^{\frac{n}{2}}}{2^{n}}=\frac{1}{2^{\frac{n}{2}}}$$

7. Oct 6, 2004

How did you even get that? You suddenly threw a proof of induction. But why did you prove it for n/2? I do not understand how you arrived at the so called "trivial" consequence.

8. Oct 6, 2004

### arildno

Do you agree that I've proven that for n>=4, we have:
$$n<=2^{\frac{n}{2}}$$?

9. Oct 6, 2004

In general, let us say we have ( n )/( a ^ n)

For the above problem wouldnt you take the square root of the sequence as such:

sqrt (n) / sqrt ( 2^n)

sqrt(2) = 1 + h

= sqrt(n) / (1+h) ^n <= sqrt(n) / (1 + nh ) <= sqrt(n) / nh

= 1 / nh^2

Could I solve for n using this method? Thank you for all your help

10. Oct 6, 2004

### arildno

Sorry I was a bit snappish earlier.
Can we focus on your first problem before generalizing?

11. Oct 6, 2004

Yeah, its ok. It's just that in my textbook I think they want me to use a certain method like the one mentioned above. I can see that your method by proof of induction works. Do you think that method mentioned above could work for solving n? Again thanks for all your input!

12. Oct 6, 2004

### arildno

Frankly, I can't make head or tails of your approach (might be cause I'm tired)

Now, there wasn't anything crucial about proving it for n/2, if that's something you might believe (you could just as easily make an induction argument on n/3 or 123/758n, for that matter).
The crucial idea is that an "exponential" grows faster than a "polynomial"
That is, an exponential function (as long as the base is greater than one) will always catch up with a polynomial. That's essentially the main idea behind the induction proof.

13. Oct 12, 2004

More Limits

Hello again. Here is the problem. In each one we are defining smaller and smaller epsilons. Can you please help? Also what if you had n^2 / 2^n = 0?

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14. Oct 12, 2004

### HallsofIvy

Staff Emeritus
At the very first, trial and error were suggested. Why did you declare "that would take too long"? It takes a heckofalot less time than reading all of these posts!

f(n)= n/2n

n f(n)
1 1/2
2 2/4= 1/2
3 3/8
4 4/16= 1/4
5 5/32= .15625
6 6/64= .09375 and we're done.

n= 6.

15. Nov 4, 2004

HallsofIvy you are correct but here is the problem:

The other problems choose epsilon smaller and smaller such as:

(n/2^n) < 1/100
(n/2^n) < 1/1000

How would I find n now? Guess and check would take a very long time. I know by the definition of convergence:

|(n/ 2^n)| < 1/100 (for example) for ever N(epsilon) > n. So when I choose epsilon smaller and smaller, n will become larger and larger. Can someone please help me?

THanks a lot

16. Nov 4, 2004

### NateTG

If you recognize that:
1. The sequence is monotone decreasing.
and
2. That it shrinks quite rapidly
guess and check is really fast.
For example, I could just try n=1, n=10, then n=100 and so on. Since the sequence is monotone, if the n'th term is inside epsilon, all the following terms are also inside epsilon, and you don't have to find the smallest n, just one that works.

Arnildo already showed that for $$n>2$$, $$\frac{2^{\frac{n}{2}}}{2^n} > \frac{n}{2^n}$$
But $$\frac{2^{\frac{2}{n}}}{2^n}=\frac{1}{2^{\frac{n}{2}}}$$
which you can easily use to get a floor for $$n_\epsilon$$

Last edited: Nov 4, 2004
17. Nov 5, 2004

yes but i need to find N1, N2, and N3 such that n> N1, n> N2, n > N3. This means that I need to find nuumbers N1, N2, and N3 such that any number greater than these numbers satisfies the inequality. Thus I cannot choose 'any number.' For example with the problem above, the answers are 6, 10, 14 respectively (1/10, 1/100, 1/1000). You notice that for each of these numbers is a lower bound. I would I find this.

Again thanks a lot everyone!

18. Nov 5, 2004