Sequences

• May 6th 2008, 09:45 AM
simplysparklers
Sequences
I'm a newbie so didn't really know where to post this, but I hope someone here can help me!I'm just beginning to study sequences, and I'm getting nowhere with it. Can anyone please help me with these two problems:

1.)Show that the sequence (1/n^k)nEN is convergent if and only if k>=0, and that the limit is 0 for all k>0.

2.)Determine the least value of N such that n/(n^2+1)<0.0001 for all n>=N
• May 6th 2008, 01:57 PM
TheEmptySet
Quote:

Originally Posted by simplysparklers
I'm a newbie so didn't really know where to post this, but I hope someone here can help me!I'm just beginning to study sequences, and I'm getting nowhere with it. Can anyone please help me with these two problems:

1.)Show that the sequence (1/n^k)nEN is convergent if and only if k>=0, and that the limit is 0 for all k>0.

2.)Determine the least value of N such that n/(n^2+1)<0.0001 for all n>=N

$p \implies q$

Since $a_n$ is convergent it must be bounded(why?)

Now consider the three cases k < 0; k=0; k> 0

if k < 0 then k = -m for some m > 0

But then $\frac{1}{n^k}=\frac{1}{n^{-m}}=n^m$
But this cannot happen because the above is not bounded.

If k=0 $a_n=1$ for all n

if k > 0 the sequence in monotically decreasing $a_{n+1}< a_{n}$ for all n

This in bounded below by zero and must converge to zero.

$q \implies p$ is fairly straight forward. use the definition

Hint: let $\epsilon > 0$ Choose $N=\left( \frac{1}{\epsilon} \right)^{1/k}$

For #2

Clear the fraction to get $n=0.0001n^2+0.0001$

Good luck.
• May 10th 2008, 05:08 AM
simplysparklers
On part 2 though, why is it $
n=0.0001n^2+0.0001
$
?and not $
n<0.0001n^2+0.0001
$
?
• May 10th 2008, 01:47 PM
TheEmptySet
Quote:

Originally Posted by simplysparklers
On part 2 though, why is it $
n=0.0001n^2+0.0001
$
?and not $
n<0.0001n^2+0.0001
$
?

It is. It is just easier to solve the equality to get n=9999.9999 but since n needs to be a natural number we choose 10,000.

P.S we don't use the other root. (why?)

I hope this helps.
• May 11th 2008, 06:23 AM
simplysparklers
It does help, thank you so much! :)

I do have to ask though, why isn't the other root used??

& in the first question, ok, I get the first part, and taking the diferent values for k, and I understand in my head because it is bounded below by 0, it must converge to 0, but I don't get the proof? Like, what do I do with
Hint: let http://www.mathhelpforum.com/math-he...fcabafdf-1.gif Choose http://www.mathhelpforum.com/math-he...69506a61-1.gif ?Sorry about this, but could you spell it out for me please?I'm never hit such a absolute immovable wall before over a maths topic, but I am not getting sequences at all (Sadsmile) and I have this assignment due in soon (Worried) Thank you so much for your time and help! (Handshake)
• May 11th 2008, 08:48 AM
TheEmptySet
Quote:

Originally Posted by simplysparklers
It does help, thank you so much! :)

I do have to ask though, why isn't the other root used??

& in the first question, ok, I get the first part, and taking the diferent values for k, and I understand in my head because it is bounded below by 0, it must converge to 0, but I don't get the proof? Like, what do I do with
Hint: let http://www.mathhelpforum.com/math-he...fcabafdf-1.gif Choose http://www.mathhelpforum.com/math-he...69506a61-1.gif ?Sorry about this, but could you spell it out for me please?I'm never hit such a absolute immovable wall before over a maths topic, but I am not getting sequences at all (Sadsmile) and I have this assignment due in soon (Worried) Thank you so much for your time and help! (Handshake)

We wish to show that $|a_n-L| < \epsilon$

let $\epsilon > 0$ be given

Let $N=\left( \frac{1}{\epsilon} \right)^{1/k}$

Then for all $n> N$ we get...

$\left( \frac{1}{\epsilon}\right)^{1/k}

moving some factors around we get

$\frac{1}{n}<(\epsilon)^{\frac{1}{k}}$

Note: the above manipulation is okay becuase n and epsilon are both positive.

Rasing both sides to the kth power we get

$\frac{1}{n^k}< \epsilon$ we will use this to prove what we want.

Now for all $n>N$ we get

$
|a_n-L|=|\frac{1}{n^k}-0|=|\frac{1}{n^k}|<\epsilon
$

QED
• May 11th 2008, 12:45 PM
simplysparklers
Quote:

Originally Posted by TheEmptySet
It is. It is just easier to solve the equality to get n=9999.9999 but since n needs to be a natural number we choose 10,000.

P.S we don't use the other root. (why?)

I hope this helps.

Thank you so much!I get all of the first one now! (Sun)
The thing I still don't get about the 2nd one,(& I'm sorry to keep bothering you on this!), is that normally you have n> [some equation with epsilon], but in his case you are just given the value of epislon, so you don't have to sub a value for epsilon into an equation to get the value of N...so where do you get the value of N?How does having n=10,000 help?

And a follow up question, you know the way it should be |a_n-L|< epsilon whenever n > N, well what if the sign is reversed?And it's |a_n-L|>= epislon for all n > N? It's in a similar question to the above one, i.e.:
Determine the least value of N such that n^2 + 2n >= 9999 for all n>N

Thank you so much for all you help!I'm stumped!
• May 11th 2008, 01:10 PM
TheEmptySet
Quote:

Originally Posted by simplysparklers
Thank you so much!I get all of the first one now! (Sun)
The thing I still don't get about the 2nd one,(& I'm sorry to keep bothering you on this!), is that normally you have n> [some equation with epsilon], but in his case you are just given the value of epislon, so you don't have to sub a value for epsilon into an equation to get the value of N...so where do you get the value of N?How does having n=10,000 help?

And a follow up question, you know the way it should be |a_n-L|< epsilon whenever n > N, well what if the sign is reversed?And it's |a_n-L|>= epislon for all n > N? It's in a similar question to the above one, i.e.:
Determine the least value of N such that n^2 + 2n >= 9999 for all n>N

Thank you so much for all you help!I'm stumped!

We are trying to find the smallest natural number N such that the inequality is true

for example with the equation $n^2+2n \ge 9999$

if we solve this for equality (and get a fraction) we can than choose the next natural number to make the inequality hold

so we want to solve
$n^2+2n=9999 \iff n^2+2n-9999=0 \iff (n+101)(n-99)=0$

so we get 2 solutions n=99 or n =-101
we get rid of -101 becuase it is not a natural number i.e (positive integer)

The above inequality will hold for all $n \ge 99$

we can check this in the original inequality

$99^2+2(99)=9801+198=9999 \ge 9999$

if you check anhy number greater than 99 it will work but any number less than 99 will not work.

I hope this clears it up.
• May 11th 2008, 01:14 PM
simplysparklers
It certainly does TheEmptySet!!Thank you so so much for your help!!You rock!!! (Rock)