# Micromass' big series challenge

• Challenge
• micromass

#### micromass

Staff Emeritus
Homework Helper
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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• QuantumQuest, ProfuselyQuarky, Andreas C and 2 others
Are 4, 6, 8 and 10 considered to be mathematical series' or infinite products?

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.

• jedishrfu
(5) doesn't look convergent. A missing minus sign?

• micromass
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?

The infinite product is defined as
$$\prod_{i=1}^{+\infty} a_i = \lim_{n\rightarrow +\infty} \prod_{i=1}^n a_i$$
Not sure what you mean with your second question.

The infinite product is defined as
$$\prod_{i=1}^{+\infty} a_i = \lim_{n\rightarrow +\infty} \prod_{i=1}^n a_i$$
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?

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.

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

• mfb, fresh_42, member 587159 and 1 other person
Solution to 6:

First we recognise that
$$\sin\left(\frac{1}{2^\alpha}\right) \prod_{k=1}^{\alpha} \cos \left(\frac{1}{2^k}\right) = \frac{1}{2^\alpha} \sin (1)$$
and hence have
$$\prod_{k=1}^{\alpha} \cos \left(\frac{1}{2^k}\right) = \frac{1}{2^{\alpha}} \frac{1}{\sin\left(\frac{1}{2^\alpha}\right)} \sin (1).$$
Now we take the limit as ##\alpha## tends to ##\infty##, noting that
$$\lim_{\alpha \to \infty} \frac{1}{2^{\alpha}} \frac{1}{\sin\left(\frac{1}{2^\alpha}\right)} \equiv \lim_{x \to 0} \frac{x}{\sin x} = 1$$
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. $$\sin\left(\frac{1}{2^{\alpha+1}}\right) \prod_{k=1}^{\alpha+1} \cos \left(\frac{1}{2^k}\right) = \frac{\sin \left(\frac{1}{2^{\alpha+1}}\right) \cos \left(\frac{1}{2^{\alpha+1}}\right)}{\sin \left(\frac{1}{2^{\alpha}}\right) } \sin\left(\frac{1}{2^\alpha}\right) \prod_{k=1}^{\alpha} \cos \left(\frac{1}{2^k}\right) = \frac{1}{2 } \sin\left(\frac{1}{2^\alpha}\right) \prod_{k=1}^{\alpha} \cos \left(\frac{1}{2^k}\right)$$
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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• mfb and micromass
First we recognise that
$$\sin\left(\frac{1}{2^\alpha}\right) \prod_{k=1}^{\alpha} \cos \left(\frac{1}{2^k}\right) = \frac{1}{2^\alpha} \sin (1)$$

How would you do this?

How would you do this?
Double angle formula.

Double angle formula.

Can you provide a somewhat longer proof for it?

(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...##

• micromass
Can you provide a somewhat longer proof for it?
Edited the original solution with a proof by induction.

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

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
$$\frac{n! e^n}{\sqrt{n}n^n}$$
converges to a value ##C##. If you can prove without using ##10## that ##C=\sqrt{2\pi}##, then I will accept your answer.

That's unfair :-)

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.

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 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.

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.

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.

• micromass
Alternative solution to Problem 10 (inspired by fresh_42's product expansion of trigonometric functions):

From wikipedia, we have the infinite product identity
$$\frac{\sin x}{x} = \prod_{k=1}^{\infty} \left(1 - \frac{x^2}{\pi^2 n^2}\right)$$
We notice that
$$\prod_{k=1}^{\infty} \left(\frac{4n^2}{4 n^2 - 1}\right) = \left[\prod_{k=1}^{\infty} \left(1 - \frac{\pi^2 / 4}{\pi^2 n^2}\right)\right]^{-1}$$
which by the above identity becomes
$$\left[\frac{\sin \left(\pi/2\right)}{\left(\pi/2\right)}\right]^{-1} = \frac{\pi}{2}$$

• mfb
Problem 7:
Not sure if this is too "cheaty", but the Taylor series expansion of ##\sin^{-1}(x)## is given by
$$\sin^{-1}(x) = \sum_{n=0}^{\infty} \frac{(2n!)}{4^{n} (n!)^2 (2n+1)}x^{2n+1}$$
The series in the problem can be rewritten as
$$\sum_{n=0}^{\infty} \frac{(2n!)}{16^{n} (n!)^2 (2n+1)} = 2 \sum_{n=0}^{\infty} \frac{(2n!)}{4^{n} (n!)^2 (2n+1)} \left(\frac{1}{2}\right)^{2n+1} = 2 \sin^{-1}(1/2)$$

• micromass
##\sum_{n=1}^{+\infty} \frac{F_{2n}}{n^2 \binom{2n}{n}}## where ##F_n## is the ##n##th Fibonacci number.

After a real Odyssey through the 18th and 19th century meeting Lucas, Catalan, Cassini, de Moivre, Binet and even old known fellows as Legendre and Jacobi, struggling with the Sirens of binomial coefficients, stealing generating functions in the cyclops' cave I finally arrived at Ithaca and made it to 9.

All my efforts have been needless - not worthless - once I found the magic sum.

##F_{2n} = \frac{1}{\sqrt{5}} \cdot (φ^{2n} - (1-φ)^{2n})## with ##φ = \frac{1+\sqrt{5}}{2}## by de Moivre's - Binet's formula. Thus
$$\sum_{n=1}^{+\infty} \frac{F_{2n}}{n^2 \binom{2n}{n}}$$
$$= \frac{2}{\sqrt{5}} \cdot \frac{1}{2} \cdot \sum_{n=1}^{+\infty} \frac{1}{n^2 \binom{2n}{n}} \cdot 2^{2n} \cdot \{ (\frac{φ}{2})^{2n} - (\frac{1-φ}{2})^{2n} \}$$
$$= \frac{2}{\sqrt{5}} \cdot (arcsin^2 (\frac{φ}{2}) - arcsin^2 (\frac{1-φ}{2}))$$
$$= 0,7062114032597...$$

Edit: Now that I have seen all the related sequences I have a nice reformulation of the sum:

##\sum_{n=1}^{+\infty} \frac{F_{2n}}{n^2 \binom{2n}{n}} = \sum_{n=1}^{+\infty} \frac{F_{n}\cdot L_n}{C_n} \cdot \frac{1}{n^3+n^2}## where ##L_n## is the Lucas sequence and ##C_n## the Catalan sequence.

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• micromass and member 587159
After a real Odyssey through the 18th and 19th century meeting Lucas, Catalan, Cassini, de Moivre, Binet and even old known fellows as Legendre and Jacobi, struggling with the Sirens of binomial coefficients, stealing generating functions in the cyclops' cave I finally arrived at Ithaca and made it to 9.

All my efforts have been needless - not worthless - once I found the magic sum.

##F_{2n} = \frac{1}{\sqrt{5}} \cdot (φ^{2n} - (1-φ)^{2n})## with ##φ = \frac{1+\sqrt{5}}{2}## by de Moivre's - Binet's formula. Thus
$$\sum_{n=1}^{+\infty} \frac{F_{2n}}{n^2 \binom{2n}{n}}$$
$$= \frac{2}{\sqrt{5}} \cdot \frac{1}{2} \cdot \sum_{n=1}^{+\infty} \frac{1}{n^2 \binom{2n}{n}} \cdot 2^{2n} \cdot \{ (\frac{φ}{2})^{2n} - (\frac{1-φ}{2})^{2n} \}$$
$$= \frac{2}{\sqrt{5}} \cdot (arcsin^2 (\frac{φ}{2}) - arcsin^2 (\frac{1-φ}{2}))$$
$$= 0,7062114032597...$$

Edit: Now that I have seen all the related sequences I have a nice reformulation of the sum:

##\sum_{n=1}^{+\infty} \frac{F_{2n}}{n^2 \binom{2n}{n}} = \sum_{n=1}^{+\infty} \frac{F_{n}\cdot L_n}{C_n} \cdot \frac{1}{n^3+n^2}## where ##L_n## is the Lucas sequence and ##C_n## the Catalan sequence.

That is correct. However, can you simplify it a bit more? Basically, I want those arcsines gone.

That is correct. However, can you simplify it a bit more? Basically, I want those arcsines gone.
Well, the result is ##\frac{4}{25\sqrt{5}}\pi^2## since the angles above are ##\frac{3}{10}\pi## and ##\frac{-1}{10}\pi##.

The deduction by series will take me a little longer and I doubt it would be a simplification.

Well, the result is ##\frac{4}{25\sqrt{5}}\pi^2## since the angles above are ##\frac{3}{10}\pi## and ##\frac{-1}{10}\pi##.

The deduction by series will take me a little longer and I doubt it would be a simplification.

Thank you, that's all I wanted to know. Nice find!

Although it cannot top Euler's identity - of course not - the formulation
$$\sum_{n\in\mathbb{N}}\frac{F_n}{n}\cdot\frac{L_n}{n}\cdot\frac{1}{(n+1)C_n} = \frac{(2\pi)^2}{\sqrt{5}^5}$$
deserves to be remembered. (I must have been to Pythagoras' school in an earlier life ... )

1. ##\sum_{n=1}^{+\infty} \frac{e^{-\frac{n}{2}}n^{n-1}}{2^{n-1} n!}##

$$\sum_{n=1}^{+\infty} \frac{e^{-\frac{n}{2}}n^{n-1}}{2^{n-1} n!} = (-2) \cdot \sum_{n=1}^{+\infty} e^{- \frac{1}{2} n} \cdot \frac{ (-n)^{n-1} }{ n! } \cdot (- \frac{1}{2} )^n = (-2) \cdot \sum_{n=1}^{+\infty} \frac{ (-n)^{n-1} }{ n! } \cdot (- \frac{1}{2\sqrt{e}} )^n = -2 W(- \frac{1}{2\sqrt{e}})$$

with Lambert's W-function, which is the inverse function of ##x → xe^x##. The sum used here is its Taylor expansion in ##0## with radius of convergence ##1/e##. Thus ## -2W(-\frac{1}{2} e^{-\frac{1}{2}}) = (-2)\cdot(-\frac{1}{2}) = 1 ## is the sum of the series.

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

Substitute $$\tan(x)=\frac{2\sin^2(x)}{\sin(2x)}$$ so the product telescopes down to
$$\frac{2^{\frac{1}{2}+\frac{1}{4}+\ldots}|\sin(1)|}{\lim_{n\to\infty}\sqrt[2^{n+1}]{|\sin(2^{n+1})|}} = 2\sin(1)\approx 1.68294196962$$
For a proof that the limit in the denominator is 1 see Agnew, R. P. and Walker, R. J. "A Trigonometric Infinite Product." Amer. Math. Monthly 54, 206-211, 1947.

• fresh_42 and mfb
Substitute $$\tan(x)=\frac{2\sin^2(x)}{\sin(2x)}$$ so the product telescopes down to
$$\frac{2^{\frac{1}{2}+\frac{1}{4}+\ldots}|\sin(1)|}{\lim_{n\to\infty}\sqrt[2^{n+1}]{|\sin(2^{n+1})|}} = 2\sin(1)\approx 1.68294196962$$
For a proof that the limit in the denominator is 1 see Agnew, R. P. and Walker, R. J. "A Trigonometric Infinite Product." Amer. Math. Monthly 54, 206-211, 1947.
How exactly do you "telescope down" with the decreasing exponents? In addition I get something like 1,68113... but my program can't run very far (≈ 450 iterations).