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Math Help - Sequences

  1. #1
    Junior Member simplysparklers's Avatar
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    Unhappy 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
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by simplysparklers View Post
    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

    Solve the quadratic for n.

    Good luck.
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  3. #3
    Junior Member simplysparklers's Avatar
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    On part 2 though, why is it <br />
n=0.0001n^2+0.0001<br />
?and not <br />
n<0.0001n^2+0.0001<br />
?
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by simplysparklers View Post
    On part 2 though, why is it <br />
n=0.0001n^2+0.0001<br />
?and not <br />
n<0.0001n^2+0.0001<br />
?
    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.
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  5. #5
    Junior Member simplysparklers's Avatar
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    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 Choose ?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 and I have this assignment due in soon Thank you so much for your time and help!
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  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by simplysparklers View Post
    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 Choose ?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 and I have this assignment due in soon Thank you so much for your time and help!

    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}<n

    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

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

    QED
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  7. #7
    Junior Member simplysparklers's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    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!
    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!
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  8. #8
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by simplysparklers View Post
    Thank you so much!I get all of the first one now!
    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.
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  9. #9
    Junior Member simplysparklers's Avatar
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    It certainly does TheEmptySet!!Thank you so so much for your help!!You rock!!!
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