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Prove that [itex]\sum_n \frac{1}{n^2} < 2[/itex]

  1. Jul 16, 2016 #1
    1. The problem statement, all variables and given/known data

    Prove that [tex] 1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}<2. [/tex]


    2. Relevant equations

    [tex] 1 = 2^0, 2 = 2^1, 3 = 2^2 - 1, 4 = 2^2, \ldots, [/tex]

    and

    [tex] 2^{n}<2^{n}+1<\ldots<2^{n+1}-1 [/tex]

    3. The attempt at a solution
    As [tex] 2^{n}<2^{n}+1<\ldots<2^{n+1}-1, [/tex] it's true that
    [tex]
    \frac{1}{2^{n}}>\frac{1}{2^{n}+1}>\ldots\frac{1}{2^{n+1}-1}.
    [/tex]
    So, being
    [tex]
    S_{n}=\left(\frac{1}{2^{n}}\right)\left(\frac{1}{2^{n}}\right)+\left(\frac{1}{2^{n}+1}\right)\left(\frac{1}{2^{n}+1}\right)+\ldots+\left(\frac{1}{2^{n+1}-1}\right)\left(\frac{1}{2^{n+1}-1}\right),
    [/tex]
    follows from previous inequalities that

    [tex] S_{n} < \left(\frac{1}{2^{n}}\right)\left(\frac{1}{2^{n}}\right)+\ldots+\left(\frac{1}{2^{n}}\right)\left(\frac{1}{2^{n}}\right) [/tex]

    [tex] =\frac{n}{2^{2n}}.
    [/tex]

    This is a dead-end to me, because I can't compare it with 2 in an elegant way. I made the graphic of this function, [tex] f(x) = x/2^{2x}, [/tex] and I verified it's less than 2 for all x, but I am pretty sure there is a better way to do this proof.
     
  2. jcsd
  3. Jul 16, 2016 #2

    Math_QED

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    Are you familiar with induction?
     
  4. Jul 16, 2016 #3
    Assume that it is greater than or equal 2 and then show that can't be true

    [tex]\sum \frac{1}{n^2}=\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\ge 2
    \\
    \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\ge 1
    \\
    \frac{2^2+3^2+4^2+...+n^2}{2^23^24^2...n^2}\ge 1
    \\[/tex]

    which you can show isn't true using a few different arguments
     
  5. Jul 16, 2016 #4

    Ray Vickson

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    [tex] \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} = \frac{2^2+3^2+4^2+...+n^2}{2^23^24^2...n^2} \:\Longleftarrow\:\text{FALSE!} [/tex]
     
  6. Jul 16, 2016 #5
    Yes, I will try this strategy.
     
  7. Jul 16, 2016 #6
    Thank you very much!
     
  8. Jul 16, 2016 #7

    haruspex

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    Your Sn only sums the 1/r2 for r from n to 2n. You need the sum all the way from 1.

    With regard to induction,you would need to set up your inductive hypothesis cleverly. Starting with "suppose the sum of the first N terms < 2" will get you nowhere. You'd need to find some other series summing to at most 2 and show that it acts as an upper bound for the given sequence.

    A useful strategy sometimes is to compare the series with the corresponding continuous function.
     
    Last edited: Jul 16, 2016
  9. Jul 16, 2016 #8
    Thanks. I will find one.
     
  10. Jul 16, 2016 #9

    rcgldr

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    I see an issue with the first of the relevant equations, based on the goal of this problem, the first relevant equation should be:

    $$ 1 = 2^0, \ 2 = 2^1, 3 = 2^1+1, \ 4 = 2^2, 5 = 2^2+1, 6 = 2^2+2, 7 = 2^2+3, 8 = 2^3, \ \ldots $$

    I don't think mathematical induction will be part of the solution. I'll wait for a reply before providing any more hints.
     
    Last edited: Jul 18, 2016
  11. Jul 17, 2016 #10

    ehild

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    Yes, it is an excellent hint!
    Compare the sum to the definite integral of 1/x2 from 1 to n.
    upload_2016-7-17_7-7-47.png
     
  12. Jul 17, 2016 #11

    haruspex

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    I was avoiding being too specific at this stage.
     
  13. Jul 17, 2016 #12

    ehild

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    Your hint "compare the series with the corresponding continuous function" was a bit misleading. I think you meant the series Σ1/n2 compared to the integral of the corresponding continuous function. With the figure, I wanted to make your hint clear .
     
  14. Jul 17, 2016 #13

    haruspex

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    Again, I was leaving it to the OP to make the analogy between the sum of the series and the integral of the function.
     
  15. Jul 17, 2016 #14

    ehild

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    Series is the sum of a sequence. https://en.wikipedia.org/wiki/Series_(mathematics). I think you meant the sum of the sequence 1/n2.
     
  16. Jul 17, 2016 #15

    haruspex

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  17. Jul 20, 2016 #16

    rcgldr

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    Combining the attempt in post #1 and the relevant equation in post #9:

    $$1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots < 1+\left(\frac{1}{2^{2}}+\frac{1}{2^{2}}\right)+\left(\frac{1}{4^{2}}+\frac{1}{4^{2}}+\frac{1}{4^{2}}+\frac{1}{4^{2}}\right)+ \ldots$$

    A finite version of the right hand series is equal to:

    $$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{N}$$

    Multiply by 2

    $$2+\frac{2}{2}+\frac{2}{4}+\frac{2}{4}+\ldots+\frac{2}{N} = 3+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{2}{N}$$

    Then subtract 2 x series - 1 x series to eliminate the inner terms:

    $$\left(3+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{2}{N}\right) - \left(1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{N}\right) = 2 - \frac{1}{N}$$

    Since

    $$1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots < 1+\left(\frac{1}{2^{2}}+\frac{1}{2^{2}}\right)+\left(\frac{1}{4^{2}}+\frac{1}{4^{2}}+\frac{1}{4^{2}}+\frac{1}{4^{2}}\right)+ \ldots$$

    Then even an infinite series of the left hand side is < 2.
     
    Last edited: Jul 20, 2016
  18. Jul 20, 2016 #17

    Ray Vickson

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    You could also use the fact that for ##n \geq 2## we have
    [tex] \frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} [/tex]
    The series
    $$\sum_{n=2}^N \left( \frac{1}{n-1} - \frac{1}{n} \right) $$
    is easy to deal with (being a so-called "telescoping" series).
     
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