Micromass' big series challenge

  • #1
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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!
 
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Answers and Replies

  • #2
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Are 4, 6, 8 and 10 considered to be mathematical series' or infinite products?
 
  • #3
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Are 4, 6, 8 and 10 considered to be mathematical series' or infinite products?
Infinite products. If you prefer to work with only series, then you can always take logarithms.
 
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  • #4
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(5) doesn't look convergent. A missing minus sign?
 
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  • #7
member 587159
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?
 
  • #8
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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.
 
  • #9
member 587159
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.
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?
 
  • #10
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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?
Yes. At least give some reasoning.
 
  • #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]
 
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  • #12
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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##.
 
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  • #13
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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]
How would you do this?
 
  • #15
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Double angle formula.
Can you provide a somewhat longer proof for it?
 
  • #16
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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...##
 
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  • #17
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Can you provide a somewhat longer proof for it?
Edited the original solution with a proof by induction.
 
  • #18
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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
 
  • #19
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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
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.
 
  • #20
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That's unfair :-)
 
  • #21
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That's unfair :-)
OK, it is. I'll mark that you solved it since I never said not to use Stirling. But anybody who can give a different proof that doesn't use Stirling (or anybody who proves Stirling) will also get credit.
 
  • #22
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I will try 1. Since it turns out to be a race and there is really a lot to type, I'll first publish my result:

##\sum_{n=1}^{+\infty} \frac{e^{-\frac{n}{2}}n^{n-1}}{2^{n-1} n!} = 2 \cdot e^{\frac{1}{2\sqrt{e}}} = 2 \exp(1/(2\sqrt{e}))≈1,354273746034562782271##
 
  • #23
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I will try 1. Since it turns out to be a race and there is really a lot to type, I'll first publish my result:

##\sum_{n=1}^{+\infty} \frac{e^{-\frac{n}{2}}n^{n-1}}{2^{n-1} n!} = 2 \cdot e^{\frac{1}{2\sqrt{e}}} = 2 \exp(1/(2\sqrt{e}))≈1,354273746034562782271##
I wouldn't bother, it's not the correct answer.
 
  • #24
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Yes, I dropped a crucial factor.

Let ##a_{n+1} = \frac{e^{-\frac{n+1}{2}}(n+1)^{n}}{2^{n} (n+1)!} = \frac{1}{n!} \cdot \frac{1}{n+1} \exp(-\frac{n}{2}) \cdot \exp(-\frac{1}{2}) \cdot (\frac{n+1}{2})^{n} = a_{n} \cdot c \cdot (\frac{n+1}{n})^{n-1}## with ##c=\frac{1}{2} e^{-\frac{1}{2}} = \frac{1}{2} \cdot a_1.##
Resolving the recursion yields ##a_{n+1} = 2 c^{n+1} \cdot \frac{(n+1)^n}{(n+1)!}##.
And here I have been blinded by the Taylor series of ##\exp(c)## and lost the factor ##(n+1)^n##.
Looks pretty divergent now.
 
  • #25
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4. The product expansion of the cosine function is ##\cos(x) = \prod_{k=1}^{k=\infty} (1 - \frac{4x^2}{(2k-1)^2 \pi^2}).##
(in old days: Leonhard Euler, nowadays: Wikipedia)
Thus ##\cos(x) = \prod_{n=0}^{n=\infty} (1 - \frac{\frac{x^2}{\pi^2}}{(n+\frac{1}{2})^2})## and ##cos(2\pi) = 1## is the value of the product in 4.
 
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