Intereting Posts

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Prove that $2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$
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I was reading about normal numbers on WikiPedia, and I ran across this statement:

Besicovitch (1935) proved that the number represented by the same

expression, with f(n) = n^2,

0.149162536496481100121144…, obtained by concatenating the square numbers in base 10, is normal in base 10.

It seems intuitively that this cannot be true, because the pattern of square numbers taken modulo 10 is `{0, 1, 4, 9, 6, 5, 6, 9, 4, 1}`

. So it seems there will be a higher distribution of those digits throughout the number. How is it possible this number could have the digits 0-9 equally represented throughout the whole decimal expansion?

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- Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$ - Part II

Because the one’s digits become an insignificant fraction of all the digits as we go along. In the part from $1^2$ through $99^2$, which is $353$ digits, $99$ of them are from the units place, or better than a quarter. When $n=10^{100}$ the ones digit contributes only one digit out of $199$ of the square and it keeps decreasing.

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