User:Iameukarya

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About Me

This user knows that 42 is the answer to the Ultimate Question of life, the Universe, and everything.

[edit] Construction

$\frac {41}{42^0} + \frac {41}{42^1} + \frac {41}{42^2} + \frac {41}{42^3} + \frac {41}{42^4} + \cdots = 42$

$\sum_{n=0}^\infty \frac {x-1}{x^n} = x$

$\frac {21}{2^1} + \frac {2(21)}{2^2} + \frac {3(21)}{2^3} + \frac {4(21)}{2^4} + \frac {5(21)}{2^5} + \cdots = 42$

$\sum_{n=1}^\infty \frac {n(\frac{x}{2})}{2^n} = x$

$\frac {41}{2} + \frac {21(41)+41}{21^1 2^2} + \frac {21^2 41 + 21(41) + 41}{21^2 2^3} + \frac {21^3 41 + 21^2 41 + 21(41) + 41}{21^3 2^4} + \cdots = 42$

$\sum_{n=0}^\infty \frac {f_n(\frac {x}{2})}{(\frac {x}{2})^n 2^{n+1}} = x$

$1 = \det(I_n)\,$

$2 = \sqrt{2}^ {\sqrt{2}^ {\sqrt{2}^ {\ \cdot^ {\cdot^ \cdot}}}} = \ ^{\infty}(\sqrt{2})$

$3 = \frac {2 \sqrt 2 \sum_{n=0}^\infty \frac{1}{16^n} \left( \frac{4}{8n + 1} - \frac{2}{8n + 4} - \frac{1}{8n + 5} - \frac{1}{8n + 6}\right)}{3267 \sum_{n=0}^\infty \frac{(4n)!(1103+26390n)}{(n!)^4 396^{4n}}\!}$

$4 = (\sqrt {i} + i \sqrt {i})^4$

$5 = \frac {( \frac {1 + \sqrt{5}}{2})^5 - ( \frac {1 - \sqrt{5}}{2})^5}{\sqrt{5}}$

$6 = \frac {3 \sqrt{(3+4+5)(3-4+5)(3+4-5)(-3+4+5)}}{4(3)}$

$7 = \ln \lim_{n \to \infty} \left (1 + \frac {7}{n}\right )^n$

$8 = \frac {16}{\pi} \prod_{n=1}^{\infty} \left( \frac{ 4 \cdot n^2 }{ 4 \cdot n^2 - 1 } \right)$

$9 = \left(\frac {9\sum_{n=1}^\infty \frac {1}{n^3}}{4 \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^3}}\right)^{\left(i \sqrt {i} - \sqrt {i}\right)\left(2 \prod_{k=0}^\infty \left(1-\frac{1}{(4k+2)^2}\right)\right)}$