# Challenge Micromass' big series challenge

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1. Jun 16, 2016

### micromass

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. Jun 16, 2016

### Staff: Mentor

Are 4, 6, 8 and 10 considered to be mathematical series' or infinite products?

3. Jun 16, 2016

### micromass

Infinite products. If you prefer to work with only series, then you can always take logarithms.

4. Jun 16, 2016

### Staff: Mentor

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

5. Jun 18, 2016

### Math_QED

6. Jun 18, 2016

### micromass

7. Jun 19, 2016

### Math_QED

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. Jun 19, 2016

### micromass

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.

9. Jun 19, 2016

### Math_QED

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. Jun 19, 2016

### micromass

Yes. At least give some reasoning.

11. Jun 19, 2016

### Svein

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: $\frac{2\sin(2)}{(2-\cos(2))^{2}+(\sin(2))^{2}}=\frac{2\sin(2)}{5-4\cos(2)}$

12. Jun 19, 2016

### Fightfish

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 $\alpha$.

Last edited: Jun 19, 2016
13. Jun 19, 2016

### micromass

How would you do this?

14. Jun 19, 2016

### Fightfish

Double angle formula.

15. Jun 19, 2016

### micromass

Can you provide a somewhat longer proof for it?

16. Jun 19, 2016

### Staff: Mentor

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

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

$= 1,0014470766409421219064785871379...$

17. Jun 19, 2016

### Fightfish

Edited the original solution with a proof by induction.

18. Jun 19, 2016

### geoffrey159

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. Jun 19, 2016

### micromass

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.

20. Jun 19, 2016

### geoffrey159

That's unfair :-)