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Help with infinite sequences and series

  1. Sep 29, 2015 #1
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    I tried the comparison test for one B but not sure if I am right. Think it could also be a ratio test because of the variable exponent. I'm lost totally lost on number one A. Also, I have the answer for the first part of three but don't know how to do the second part of it by comparing.

    Thanks
     
  2. jcsd
  3. Sep 29, 2015 #2

    HallsofIvy

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    For A, do you know that [itex]1+ 2+ 3+ \cdot\cdot\cdot + k= \frac{k(k+1)}{2}[/itex]?
    For B, [itex]\sqrt{k}- \sqrt{k+1}= \frac{(\sqrt{k}- \sqrt{k+1})(\sqrt{k}+ \sqrt{k+ 1}}{\sqrt{k}+ \sqrt{k+1}}= \frac{-1}{\sqrt{k}+ \sqrt{k+1}}[/itex]

    For 3, you say you have already determined the values of p such that the given series converges. Okay, for what values of p does [itex]\sum_{n= 1}^\infty \frac{1}{n^p}[/itex] converge?
     
    Last edited by a moderator: Sep 29, 2015
  4. Sep 29, 2015 #3

    Krylov

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    Hint: For 1(a) first recall the formula for the first ##k## natural numbers, then use partial fraction decomposition. After that, you will see lots of things cancelling.
     
  5. Sep 29, 2015 #4

    Ray Vickson

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    Do you know how to evaluate the sum ##1+2+3 + \cdots + k##? I cannot believe you have not seen it before, and if not, try a Google search.
     
  6. Sep 29, 2015 #5
    Ray, isn't the limit of k the summation? So, it would be infinity? I just finished reading sequences and series but prof. is already 3 sections ahead. Stewart's Calculus isn't helping so these instructional google videos (e.g. Khan Academy and prof. Leonard) slow me down. Halls of ivy, how did you get rid of the variable exponent? Krylov, I'm not really good at partial fraction decomposition and the only formula I know is two from geometric series and just bunch of theorems at the moment.
     
  7. Sep 29, 2015 #6

    Ray Vickson

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    In 1(a) your sum has the form
    [tex] S = \frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3} + \cdots + \frac{1}{1+2+ \cdots + k} + \cdots [/tex]
    I am asking you if you can find a formula for the ##k##th term of the sum; later on, you can worry about whether the sum is convergent. But first, you need to get the kth term in a more manageable form.

    As I said, if you do not know the sum in the denominator of term k, you can go to the library and look it up, or nowadays do an on-line search.
     
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