Challenge Micromass' big series challenge

[micromass]

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!

 

[fresh_42]

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 ...)

 

[fresh_42]

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.

 

[bpet]

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]

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).

 

[mfb]

The root increases by a factor 2, which is exactly the factor 2 you have in the numerator (sin²) vs. the denominator.

It is quite interesting that 2ⁿ⁺¹ does not get too often too close to a zero of the sine to ruin the limit.