User:Iameukarya

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

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

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

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

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

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

$$1 = \det(I_n)\,$$

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

$$3 = \frac {\sqrt 2 \sum_{n=0}^\infty \frac{(4n)!(1103+26390n)}{(n!)^4 396^{4n}}\!}{19602 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 640320^{3k + 3/2}}}$$

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

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

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

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

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

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

$$10 = \prod_{n=1}^{\infty} e^{\frac {(\frac {9}{10})^n}{n}}$$

$$0 = \frac {\pi}{\ln -1} + i$$

$$x = \prod_{n=1}^{\infty} e^{\frac {(\frac {x-1}{x})^n}{n}}$$

$$x^2 = -1 \,$$

$$x^2 - 1 = -2 \,$$

$$(x - 1)(x + 1) = -2 \,$$

$$x - 1 = \frac {-2}{1 + x}$$

$$x = 1 - \frac {2}{1 + x}$$

$$x = 1 - \frac {2}{1 + \Big(1 - \cfrac {2}{1 + x}\Big)}$$

$$x = 1 - \frac {2}{2 - \cfrac {2}{1 + x}}$$

$$x = 1 - \frac {2}{2 - \cfrac {2}{2 - \cfrac {2}{2 - \cfrac {2}{2 - \cfrac {2}{\ddots}}}}}$$

$$e^{i \pi}=-1 \,$$

$$n = \ ^{\infty}\Big(n^{n^{-1}}\Big)$$

$$\sin x = \frac {opposite}{hypotenuse}$$

$$\cos x = \frac {hypotenuse}{opposite}$$

$$\sec x = \frac {adjacent}{hypotenuse}$$

$$\csc x = \frac {hypotenuse}{adjacent}$$

$$\tan x = \frac {\sin x}{\sec x} = \frac {\csc x}{\cos x}$$

$$\cot x = \frac {\sec x}{\sin x} = \frac {\cos x}{\csc x}$$

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

$$\lim_{n \to \infty}	n\left ( x^\frac{1}{n} - 1 \right )$$

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

$$\left ( 1 - \frac{1}{2^{s-1}} \right )\zeta (s) = \eta (s)$$

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

$$\lim_{n \to \infty}	\left ( 1 - \frac{1}{2^{1/n}} \right )n = \ln 2$$

$$\lim_{n \to \infty}	n \left ( 1 - \frac{1}{x^{1/n}} \right )$$

$$\sum_{n=1}^\infty \left (\sum_{m=1}^{x-1} \frac{1}{xn - m} - \frac{x-1}{xn}\right ) = \ln x$$

$$\lim_{x \to \infty} \sum_{n=2}^\infty \left (\sum_{m=1}^{x-1} \frac{1}{xn - m} - \frac{x-1}{xn}\right ) = 1 - \gamma$$

$$\sum_{n=1}^\infty \left (\frac{1}{xn}\sum_{m=1}^{x-1} \frac{m}{xn - m} \right ) = \ln x$$

$$1 - \frac{1}{3^n} + \frac{1}{5^n} - \frac{1}{7^n} + \frac{1}{9^n} - \frac{1}{11^n} \pm \cdots$$

$$\frac{1}{3} - \frac{1-\tfrac{1}{3}}{5} + \frac{1-\tfrac{1}{3}+\tfrac{1}{5}}{7} - \frac{1-\tfrac{1}{3}+\tfrac{1}{5}-\tfrac{1}{7}}{9} \pm \cdots = \frac{\pi^2}{32}$$

$$\prod_{p}^\infty \left (1 - \frac{1}{{p}^n}\right)^{-1} = \zeta(n)$$

$$\prod_{p \neq 2}^\infty \left (1 - \frac{i^{(p - 1)}}{{p}^n}\right)^{-1}$$

$$\prod_{p}^\infty \left (1 + \frac{1}{{p}^n}\right)^{-1} \prod_{q}^\infty \left (1 - \frac{1}{q^n}\right)^{-1}$$

$$\sum_{k=0}^\infty \frac{1}{(2k+1)^n}=\left(1 - \frac{1}{2^n}\right) \zeta(n)$$

$$\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^n}=\left(1 - \frac{1}{2^n}\right) \zeta(n) \prod_{p}^\infty \left(\frac{p^n - 1}{p^n + 1}\right)$$

$$\frac{\pi}{2} = \sum_{n=1}^\infty \frac{(-1|n)}{n}$$

$$\prod_{p}^\infty \left(\frac{p^n - 1}{p^n + 1}\right) \zeta(n) = \sum_{k=1}^\infty \frac{(-1|k)}{k^n}$$

$$\frac{\pi}{2} = 1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} - \frac{1}{11} - \frac{1}{12} + \frac{1}{13} - \frac{1}{14} - \frac{1}{15} + \frac{1}{16} + \cdots$$

$$\frac{1}{4} = \sum_{n=0}^\infty \left (1-\beta(2n+1)\right)$$

$$\ln \sqrt{2} = \sum_{n=1}^\infty \left (1-\beta(n)\right)$$

$$\beta(2n+1) = \frac{E_k \pi^{2n+1}}{4(4n)!!}$$

$$\frac{\pi}{4} = \sum_{n=1}^\infty \frac{\eta (n)}{2^n}$$

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

$$1 + 1 + 1 + 1 + 1 + \cdots = -\frac{1}{2}$$

$$1 + 2 + 3 + 4 + 5 + \cdots = -\frac{1}{12}$$

$$1 + 4 + 9 + 16 + 25 + \cdots = 0$$

$$1 + 8 + 27 + 64 + 125 + \cdots = \frac{1}{120}$$

$$1 + 16 + 81 + 256 + 625 + \cdots = 0$$

$$1 + 32 + 243 + 1024 + 3125 + \cdots = -\frac{1}{252}$$

$$1 \times 2 \times 3 \times 4 \times 5 \times \cdots = \sqrt{2\pi}$$

$$n(r) = \frac{1}{3}\left(\frac{r^2}{(r^6 + 54a^2r^2 + 6 \sqrt{3}\sqrt{27a^4r^8 + a^2r^{10}})^\frac{1}{3}} + \frac{(r^6 + 54a^2r^2 + 6 \sqrt{3}\sqrt{27a^4r^8 + a^2r^{10}})^\frac{1}{3}}{r^2} - 2 \right)$$