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P-series converges?

  1. Mar 5, 2016 #1
    1. The problem statement, all variables and given/known data
    http://math.oregonstate.edu/home/pr...stStudyGuides/SandS/SeriesTests/p-series.html

    2. Relevant equations


    3. The attempt at a solution
    why when the p is between 0 and 1 , the p-series diverges? when p is between 0 and 1 , the denominator still become big , when 1/ big number , the number will become smaller than before. ,So, the series will converge , right?
     
  2. jcsd
  3. Mar 5, 2016 #2

    stevendaryl

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    No, just because the terms get smaller and smaller doesn't mean that the series converges. They give the example of the series [itex]1 + \frac{1}{2} + \frac{1}{3} + ...[/itex], which is a p-series with [itex]p=1[/itex]. That doesn't converge. To see that it doesn't, we can group the terms this way:

    [itex]S = 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + ...[/itex]

    (In each group, you have a sum of the form [itex]\frac{1}{2^n + 1} + ... + \frac{1}{2^n + 2^n}[/itex])

    If S converges, then certainly it would still converge if you replaced terms by smaller terms. So in each group, replace each term by the smallest term in the group:
    [itex]S' = 1 + \frac{1}{2} + (\frac{1}{4} + \frac{1}{4}) + (\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}) + ...[/itex]

    If [itex]S[/itex] converged, then so would [itex]S'[/itex]. But [itex]S' = 1 + 1/2 + 1/2 + 1/2 + ...[/itex] which clearly diverges.

    If you choose [itex]p < 1[/itex], then the series diverges even worse.
     
  4. Mar 5, 2016 #3
    You can see that it diverges when ##0<p<1 ## by comparison to the harmonic serie: you have ##\lim_{n\to\infty} n\times \frac{1}{n^p} = +\infty##, and therefore ##\exists N\in \mathbb{N}, \ (n > N \Rightarrow \frac{1}{n^p} > \frac{1}{n}) ##.
     
  5. Mar 5, 2016 #4

    Ray Vickson

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    Note that for ##p>0## the function ##1/x^p## is strictly decreasing, so
    [tex] \int_n^{n+1} 1/x^p \: dx < 1/n^p < \int_{n-1}^n 1/x^p \: dx [/tex]
    Summing over ##n \leq N## we have
    [tex] \int_1^{N+1} x^{-p} \, dx < \sum_{n=1}^N 1/n^p < 1 + \int_1^N x^{-p} \, dx [/tex]
    The integrals are both do-able, and for ##0 < p \leq 1## we have ##\sum 1/n^p \geq \int_1^{\infty} x^{-p} \, dx = +\infty##, so the sum is divergent. If ##p > 1## the partial sums are bounded above by ##1+\int_1^{\infty} x^{-p} \, dx = p/(p-1)##, so ##\sum 1/n^p## converges, and its value lies between 1 and ##p/(p-1)##.
     
    Last edited: Mar 5, 2016
  6. Mar 5, 2016 #5
    i don't under stand, can you explain further?
     
  7. Mar 5, 2016 #6
    If you take a positive terms serie, it either converges or diverges to ##+\infty##. In the case of the harmonic serie, it was explained in first post why it diverges ( to ##+\infty## ). From the inequality above, it is clear that the serie of ##\{1/n^p\}_n## will also diverge to ##+\infty## (when ## p < 1 ##)
     
  8. Mar 5, 2016 #7
    i dont understand why the harmonic series diverges? can you explain ?
     
  9. Mar 6, 2016 #8

    Ray Vickson

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    Read reply #4; it is all in there.
     
  10. Mar 6, 2016 #9
    can you explain on it ? i dont understand
     
  11. Mar 6, 2016 #10

    Ray Vickson

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    [tex] 1 + \int_1^N x^{-p} \, dx < 1 + \int_1^{\infty} x^{-p} \, dx, [/tex]
    because ##x^{-p} > 0## for all ##x > 0## and
    [tex] \int_1^{\infty} x^{-p} dx = \int_1^N x^{-p} dx +
    \underbrace{\int_N^{\infty} x^{-p} dx}_{>0} [/tex]
     
    Last edited: Mar 6, 2016
  12. Mar 6, 2016 #11
    can you explain why there is 1 appear in the second line ?
     
  13. Mar 6, 2016 #12
    why it will equals to p / p-1 ?
     
  14. Mar 6, 2016 #13

    Ray Vickson

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    Do the integral and see what you get.
     
  15. Mar 6, 2016 #14
    i got this
     

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  16. Mar 6, 2016 #15
    Mr Vickson answers your question "why the harmonic serie diverges?" by comparing the harmonic serie to an integral. The conclusion of it is that the partial sum ##H_N = \sum_{n=1}^N \frac{1}{n}## is equivalent as ##N \to +\infty## to ##\ln N##. This means that as ##N\to \infty##, ##H_N / \ln N \to 1 ##, showing that ##H_N## can't have a finite limit.
     
  17. Mar 6, 2016 #16

    stevendaryl

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    Several people (including me) have already explained. Just look at the partial sums:
    1. The first term is greater than [itex]\frac{1}{2}[/itex]
    2. The sum of the first two terms is greater than [itex]\frac{2}{2}[/itex]
    3. The sum of the first four terms is greater than [itex]\frac{3}{2}[/itex]
    4. The sum of the first eight terms is greater than [itex]\frac{4}{2}[/itex]
    5. In general, if you sum up the first [itex]2^n[/itex] terms, you get something greater than [itex]\frac{n+1}{2}[/itex]
    So the partial sums keep growing bigger and bigger, without bound. For a series to be convergent, the partial sums have to be bounded.
     
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