Finding Limits of Sequences: How to Determine N Values for Given Conditions

[courtrigrad]

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

 

[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.

 

[courtrigrad]

i am looking for the minimum N_1


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

Thanks

 

[learningphysics]

i am looking for the minimum N_1

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?

 

[courtrigrad]

any other ideas?

thanks

 

[courtrigrad]

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

 

[courtrigrad]

Thanks anyway for your help

 

[courtrigrad]

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 &lt; \frac {1}{10} for every n &gt; N_1
\sqrt n+1 - \sqrt n &lt; \frac {1}{100} for every n &gt; N_2
\sqrt n+1 - \sqrt n &lt; \frac {1}{1000} for every n &gt; 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}<br /> <br /> How would i obtain N_1, N_2 ,N_3?<br /> <br /> PS: I know if I solve the equation I get <b> a </b> N_1 N_2 N_3 but I want to get the least of all of them<br /> <br /> Any help is appreciated (as with the first one do I just guess and check?)<br /> <br /> Thanks

 

[courtrigrad]

hmmm, i i think i have to substitute values in lless than the solved value?

thanks