# Limits of Sequences

1. Jan 21, 2005

Hello all:

Given $$a_n = \frac {n}{\alpha^n}$$ and $$\alpha$$ is a number greater than 1, we assert as n increases the sequence of numbers $$a_n = \frac {n}{\alpha^n}$$ tends to the limit 0.

Let us consider the sequence $$\sqrt a_n = \frac {\sqrt n}{(\sqrt \alpha)^n}$$

We put $$\sqrt \alpha = 1+h$$ where $$h > 0$$

$$\sqrt \alpha^n = (1+h)^n > 1+nh$$

$$\sqrt a_n = \frac {\sqrt n}{(1+h)^n} \leq \frac {\sqrt n}{1+nh} \leq \frac {\sqrt n}{nh} = \frac {1}{h\sqrt n}$$

$$a_n \leq \frac {1}{nh^2}$$ and $$a_n \rightarrow 0$$

Now my question is: Suppose you are asked to find numbers $$N_1, N_2, N_3$$ such that

(a) $$\frac {n}{2^n} < \frac {1}{10}$$ for every $$n > N_1$$
(b) $$\frac {n}{2^n} < \frac {1}{100}$$ for every $$n > N_2$$
(c) $$\frac {n}{2^n} < \frac {1}{1000}$$ for every $$n > N_3$$

Do I have to guess and check for these numbers or can I somehow use the proof of this sequence? Because what if $$\epsilon$$ becomes smaller and smaller? Then there must be some method to find $$N_1, N_2, N_3$$

Thanks a lot for any help or advice

EDIT: Its $$\frac {n}{2^n} < \frac {1}{10}$$ for every $$n > N_1$$ and the same with the others

Last edited: Jan 21, 2005
2. Jan 21, 2005

### learningphysics

In your case here: $$\alpha = 2$$ Solve for h.

You want $$a_n<\frac{1}{10}$$. If you know that $$\frac{1}{nh^2}<\frac{1}{10}$$ then that means that $$a_n<\frac{1}{10}$$.

So from $$\frac{1}{nh^2}<\frac{1}{10}$$ you can get the n value you need. This will give you an N1 such that if n>N1, $$a_n<\frac{1}{10}$$

But are you looking for the minimum N1 such that if n>N1 then $$a_n<\frac{1}{10}$$, or are you just looking for any N1 such that if n>N1 then $$a_n<\frac{1}{10}$$? If you're looking for the minimum N1, then you might want to want to calculate $$a_{N1}$$ and make sure that it is >=1/10.

3. Jan 21, 2005

i am looking for the minimum $$N_1$$

I know that for part (a) $$N_1 = 6$$ buy why?

Thanks

Last edited: Jan 21, 2005
4. Jan 21, 2005

### learningphysics

Then I think you'd just have to plug it in and check.

Are you trying to do an epsilon-delta proof to show that the limit is 0? Then it is sufficient to show that for every epsilon>0, there is some N such that for n>N |a(n)-0|<epsilon or in other words show that a(n)<epsilon. I'm guess that the 1/10, 1/100 are different choices for epsilon?

5. Jan 21, 2005

any other ideas?

thanks

6. Jan 21, 2005

so i take it that the only way to find the smallest numbers would be by guess and check? Because after solving for n i get 64.

This is from Courant's book by the way (i am a high school student)

Thanks

Last edited: Jan 22, 2005
7. Jan 22, 2005

8. Jan 22, 2005

Same concept Different Limit

Ok so let's say we have $$a_n = \sqrt n+1 - \sqrt n$$ and we want to find numbers $$N_1, N_2, N_3$$ such that

$$\sqrt n+1 - \sqrt n < \frac {1}{10}$$ for every $$n > N_1$$
$$\sqrt n+1 - \sqrt n < \frac {1}{100}$$ for every $$n > N_2$$
$$\sqrt n+1 - \sqrt n < \frac {1}{1000}$$ for every $$n > N_3$$

Now since $$\sqrt n+1 - \sqrt n = \frac {(\sqrt n+1 - \sqrt n)(\sqrt n+1 + \sqrt n)}{\sqrt n+1 + \sqrt n} = \frac {1}{\sqrt n+1 + \sqrt n} How would i obtain [tex] N_1, N_2 ,N_3$$?

PS: I know if I solve the equation I get a $$N_1 N_2 N_3$$ but I want to get the least of all of them

Any help is appreciated (as with the first one do I just guess and check?)

Thanks

9. Jan 22, 2005