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Challenge Micromass' big series challenge

  1. Jun 16, 2016 #1

    micromass

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    We had integrals, so we have to have series as well. Here are 10 easy to difficult series and infinite products. Up to you to find out the exact sum.

    Rules:
    • The answer must be a finite expression. The only expressions allowed are integers written in base 10, the elementary arithmetic operations, trigonometric functions and their inverses, exponential functions and their inverses, ##\pi## and imaginary numbers.
    • Any use of sources is allowed as long as you don't search for the question directly. So you can search a calculus book, but not type in the question in wolframalpha. Any use of sources must be detailed in the answer.
    • The first person to give a complete answer will get the recognition. Sometimes another person will be given the recognition as well.
    • If you encountered the question before and remember the answer, you cannot participate in this question.
    Problems:
    1. SOLVED BY fresh_42 ##\sum_{n=1}^{+\infty} \frac{e^{-\frac{n}{2}}n^{n-1}}{2^{n-1} n!} = 1##

    2. SOLVED BY Math_QED ##\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{4n^2 - 1} = \frac{\pi-2}{4}##

    3. SOLVED BY fresh_42 ##\sum_{n=0}^{+\infty} \frac{1}{(2n + 1)^6} = \frac{\pi^6}{960}##

    4. SOLVED BY fresh_42 ##\prod_{n=0}^{+\infty} \left(1 - \frac{4}{\left(n + \frac{1}{2}\right)^2}\right) = 1##

    5. SOLVED BY Svein ##\sum_{n=1}^{+\infty} 2^{-n} \sin(2n) = \frac{2\sin(2)}{5-4\cos(2)}##

    6. SOLVED BY Fightfish ##\prod_{k=1}^{+\infty} \cos\left(\frac{1}{2^k}\right) = \sin(1)##

    7. SOLVED BY Fightfish ##\sum_{n=0}^{+\infty} \frac{(2n)!}{16^n (n!)^2 (2n+1)}=2\sin^{-1}\left(\frac{1}{2}\right)##

    8. SOLVED BY bpet ##\prod_{n=0}^{+\infty} \sqrt[2^{n+1}]{|\tan(2^n)|}##

    9. SOLVED BY fresh_42 ##\sum_{n=1}^{+\infty} \frac{F_{2n}}{n^2 \binom{2n}{n}}=\frac{4}{25\sqrt{5}}\pi^2## where ##F_n## is the ##n##th Fibonacci number.

    10. SOLVED BY geoffrey159 ##\prod_{n=1}^{+\infty} \frac{4n^2}{4n^2 - 1} = \frac{\pi}{2}##
    Thank you for participating, I hope you have fun! Comments welcome!
     
    Last edited: Jul 16, 2016
  2. jcsd
  3. Jun 16, 2016 #2

    jedishrfu

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    Are 4, 6, 8 and 10 considered to be mathematical series' or infinite products?
     
  4. Jun 16, 2016 #3

    micromass

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    Infinite products. If you prefer to work with only series, then you can always take logarithms.
     
  5. Jun 16, 2016 #4

    mfb

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    (5) doesn't look convergent. A missing minus sign?
     
  6. Jun 18, 2016 #5

    Math_QED

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  7. Jun 18, 2016 #6

    micromass

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  8. Jun 19, 2016 #7

    Math_QED

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    2 quick questions, as I am not familiar with infinite products.

    1) Is the limit k-> infinity from the kth partial sum equal to the infinite product?
    2) If I would have found a partial product, do I need to show this is the real partial product?
     
  9. Jun 19, 2016 #8

    micromass

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    The infinite product is defined as
    [tex]\prod_{i=1}^{+\infty} a_i = \lim_{n\rightarrow +\infty} \prod_{i=1}^n a_i[/tex]
    Not sure what you mean with your second question.
     
  10. Jun 19, 2016 #9

    Math_QED

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    Thanks

    I mean, if I would have found a partial product by writing out the product of the n first factors and see a patern, should I use induction or something else to show that this is always true?
     
  11. Jun 19, 2016 #10

    micromass

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    Yes. At least give some reasoning.
     
  12. Jun 19, 2016 #11

    Svein

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    As to 5:
    Consider the sum [itex]\sum_{n=1}^{\infty}2^{-n}e^{2ni} [/itex]. The imaginary part of this sum is what we are looking for. So, let [itex]a_{n}=2^{-n}e^{2ni}=e^{-n\ln(2)+2ni}=e^{-n(\ln(2)-2i)} [/itex]. Then [itex]\frac{a_{n+1}}{a_{n}}=\frac{e^{-(n+1)(\ln(2)-2i)}}{e^{-n(\ln(2)-2i)}}=e^{-(\ln(2)-2i)}=\frac{1}{2}e^{2i} [/itex]. Since this ratio is <1, we can use the formula for the sum of a geometric sequence: [itex] s=\frac{1}{1-k}=\frac{1}{1-\frac{1}{2}e^{2i}}=\frac{2}{2-e^{2i}}[/itex]. We need the imaginary part of this expression, so transform it: [itex] \frac{2}{2-e^{2i}}=\frac{2}{2-(\cos(2)+i\sin(2))}=\frac{2}{(2-\cos(2))-i\sin(2)}=\frac{2((2-\cos(2))+i\sin(2))}{(2-\cos(2))^{2}+(\sin(2))^{2}}[/itex]. The imaginary part of this is [itex]\frac{2 \sin(2)}{(2-\cos(2))^{2}+(\sin(2))^{2}} [/itex]. This expression can be simplified somewhat: [itex]\frac{2\sin(2)}{(2-\cos(2))^{2}+(\sin(2))^{2}}=\frac{2\sin(2)}{5-4\cos(2)} [/itex]
     
  13. Jun 19, 2016 #12
    Solution to 6:

    First we recognise that
    [tex]\sin\left(\frac{1}{2^\alpha}\right) \prod_{k=1}^{\alpha} \cos \left(\frac{1}{2^k}\right) = \frac{1}{2^\alpha} \sin (1)[/tex]
    and hence have
    [tex]\prod_{k=1}^{\alpha} \cos \left(\frac{1}{2^k}\right) = \frac{1}{2^{\alpha}} \frac{1}{\sin\left(\frac{1}{2^\alpha}\right)} \sin (1).[/tex]
    Now we take the limit as ##\alpha## tends to ##\infty##, noting that
    [tex]\lim_{\alpha \to \infty} \frac{1}{2^{\alpha}} \frac{1}{\sin\left(\frac{1}{2^\alpha}\right)} \equiv \lim_{x \to 0} \frac{x}{\sin x} = 1[/tex]
    and hence the answer is ##\sin(1)##.

    Edit: micromass requested for an extended proof of the first equation, so here it is:
    1. Iterated double angle formulas
    2. [tex]\sin\left(\frac{1}{2^{\alpha+1}}\right) \prod_{k=1}^{\alpha+1} \cos \left(\frac{1}{2^k}\right)
    = \frac{\sin \left(\frac{1}{2^{\alpha+1}}\right) \cos \left(\frac{1}{2^{\alpha+1}}\right)}{\sin \left(\frac{1}{2^{\alpha}}\right) }
    \sin\left(\frac{1}{2^\alpha}\right) \prod_{k=1}^{\alpha} \cos \left(\frac{1}{2^k}\right)
    = \frac{1}{2 } \sin\left(\frac{1}{2^\alpha}\right) \prod_{k=1}^{\alpha} \cos \left(\frac{1}{2^k}\right)
    [/tex]
    Evidently, if the equation is true for some ##\alpha##, then it is also true for ##\alpha+1##. The base case ##\alpha = 1## is clearly true, so by induction, this holds for all integer ##\alpha##.
     
    Last edited: Jun 19, 2016
  14. Jun 19, 2016 #13

    micromass

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    How would you do this?
     
  15. Jun 19, 2016 #14
    Double angle formula.
     
  16. Jun 19, 2016 #15

    micromass

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    Can you provide a somewhat longer proof for it?
     
  17. Jun 19, 2016 #16

    fresh_42

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    (3) I'm not sure whether this is already cheating, but ##\zeta(6) = \frac{\pi^6}{945}##.

    Therefore ##\sum_{n=0}^{+\infty} \frac{1}{(2n + 1)^6} = \zeta(6) - \sum_{n=0}^{+\infty} \frac{1}{(2n)^6} = \zeta(6) - \frac{1}{64} \zeta(6) = \frac{63}{64\cdot 945} \pi^6 = \frac{\pi^6}{960}##

    ## = 1,0014470766409421219064785871379...##
     
  18. Jun 19, 2016 #17
    Edited the original solution with a proof by induction.
     
  19. Jun 19, 2016 #18
    Number 10, the infinite product is finite and equal to ##\pi / 2##. For this I wrote

    ##\prod_{n = 1 }^N \frac{4n^2}{4n^2-1} = 4^N (N!)^2 \frac{2^N N!}{(2N)!} \frac{2^N N!}{(2N+1)!} = \frac{4^{2N} (N!)^4}{(2N+1) ((2N)!)^2 } ##

    And Stirling's formula leads to the conclusion
     
  20. Jun 19, 2016 #19

    micromass

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    The problem is that many proofs of Stirling's formula use exactly this infinite series. I will accept that
    [tex]\frac{n! e^n}{\sqrt{n}n^n}[/tex]
    converges to a value ##C##. If you can prove without using ##10## that ##C=\sqrt{2\pi}##, then I will accept your answer.
     
  21. Jun 19, 2016 #20
    That's unfair :-)
     
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