$$\eqalign{ & s > 0,{\text{ }}s \ne \frac{1}{n}; \cr & \cfrac{1}{{\cfrac{s}{{1 + \cfrac{{s - 1}}{{1 + \cfrac{{2s}}{{1 + \cfrac{{2s - 1}}{{1 + \cfrac{{3s}}{{1 + \cfrac{{3s - 1}}{{1 + ...}}}}}}}}}}}}}} = {s^{\frac{1}{s} - 1}} \cdot {e^{\frac{1}{s}}} \cdot \Gamma \left( {\frac{1}{s},\frac{1}{s}} \right) \cr} $$

\[\coth x = \frac{{{e^{2x}} + 1}}{{{e^{2x}} - 1}} = \frac{1}{x} + \frac{{\frac{1}{{1 \cdot 3}}}}{{\frac{1}{x} + \frac{{\frac{1}{{3 \cdot 5}}}}{{\frac{1}{x} + \frac{{\frac{1}{{5 \cdot 7}}}}{{\frac{1}{x} + ...}}}}}}.\]

\[\tanh \frac{1}{n} = \frac{1}{{n + \frac{1}{{3n + \frac{1}{{5n + \frac{1}{{7n + \frac{1}{{...}}}}}}}}}}\]

$$\eqalign{ & \cfrac{1}{{1 + x}} + \cfrac{{\cfrac{1}{{1 + x/2}} + \cfrac{{\cfrac{1}{{1 + x/3}} + \cfrac{{...}}{{3/x}}}}{{2/x}}}}{{1/x}} = \cr & = x \cdot {\left( { - x} \right)^{ - x}} \cdot \left( {\Gamma \left( x \right) - \Gamma \left( {x, - x} \right)} \right) \cr & = x \cdot {\left( { - x} \right)^{ - x}}\int\limits_0^{ - x} {{t^{x - 1}}{e^{ - t}}dt} = x{\left( { - x} \right)^{ - x}}\gamma \left( {x, - x} \right) \cr} $$

$$1 + \cfrac{{1 + \cfrac{{1 + \cfrac{{1 + \cfrac{{1 + ...}}{{4s + 1}}}}{{3s + 1}}}}{{2s + 1}}}}{{s + 1}} + \cfrac{1}{{\cfrac{s}{{1 + \cfrac{{s - 1}}{{1 + \cfrac{{2s}}{{1 + \cfrac{{2s - 1}}{{1 + \cfrac{{3s}}{{1 + \cfrac{{3s - 1}}{{1 + ...}}}}}}}}}}}}}} = {s^{\frac{1}{s} - 1}}{e^{\frac{1}{s}}}\Gamma \left( {\frac{1}{s}} \right)$$ \[s > 0,{\text{ }}s \ne \frac{1}{n};\]

$$1 + \cfrac{{\cfrac{1}{{1 \cdot s}}}}{{1 + \cfrac{{\cfrac{1}{{2 \cdot s}}}}{{1 + \cfrac{{\cfrac{1}{{3 \cdot s}}}}{{1 + ...}}}}}} = 1 + \cfrac{{1 + \cfrac{{1 + \cfrac{{1 + ...}}{{3s + 1}}}}{{2s + 1}}}}{{1s + 1}} = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{\prod\limits_{k = 1}^n {\left( {sk + 1} \right)} }}} = {s^{\frac{1}{s} - 1}}{e^{\frac{1}{s}}}\left( {\Gamma \left( {\frac{1}{s}} \right) - \Gamma \left( {\frac{1}{s},\frac{1}{s}} \right)} \right)$$

Найдите на картинке все группы изоморфных графов.

$$\begin{array}{l} {a_1} = 0;{a_2} = 1; \hfill \\ {a_n} = \sqrt {\frac{{{a_{n - 1}} + {a_{n - 2}}}}{2} \cdot {a_{n - 1}}} ,{\text{ }}n \geqslant 3; \hfill \\ \mathop {\lim }\limits_{n \to \infty } {a_n} = \frac{1}{\pi }{\rm B}\left( {\frac{1}{2},\frac{3}{4}} \right) \hfill \\ \end{array} $$

$$\prod\limits_{k = 1}^{ + \infty } {\frac{{{{\left( {3k + 2} \right)}^3}}}{{27k{{\left( {k + 1} \right)}^2}}}} = {\left( {\frac{3}{{2\Gamma \left( {\frac{2}{3}} \right)}}} \right)^3}$$

$$\prod\limits_{k = 1}^{ + \infty } {\frac{{2k\left( {2k + 2} \right)}}{{{{\left( {2k + 1} \right)}^2}}}} = \frac{\pi }{4}$$

$$\eqalign{ & \prod\limits_{k = 1}^{ + \infty } {{{\left( {1 + \frac{1}{{2k}}} \right)}^{{{\left( { - 1} \right)}^k}}}} = \frac{4}{{{\rm B}\left( {\frac{1}{2},\frac{1}{4}} \right)}} \cr & \prod\limits_{k = 1}^{ + \infty } {{{\left( {1 - \frac{1}{{2k}}} \right)}^{{{\left( { - 1} \right)}^k}}}} = \frac{1}{\pi }{\rm B}\left( {\frac{1}{2},\frac{1}{4}} \right) \cr & \prod\limits_{k = 1}^{ + \infty } {\left( {1 + \frac{{{{\left( { - 1} \right)}^{k + 1}}}}{{2k}}} \right)} = \frac{1}{{2\pi }}{\rm B}\left( {\frac{1}{4},\frac{1}{4}} \right) \cr} $$

$$\int\limits_0^{ + \infty } {\operatorname{arctg} \left( {{e^{ - x}}} \right)dx} $$

$$\sum\limits_{k = 0}^{ + \infty } {\frac{1}{{\cosh \left( {\pi k} \right)}}} = \frac{1}{2} + \frac{1}{{{\rm B}\left( {\frac{3}{4},\frac{3}{4}} \right)}}.$$ $$\sum\limits_{k = 0}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^k}}}{{\cosh \left( {\pi k} \right)}}} = \frac{1}{2} + \frac{1}{{{\rm B}\left( {\frac{1}{2},\frac{3}{4}} \right)}}.$$

$$\int\limits_0^{ + \infty } {\frac{1}{{\sqrt {{e^x} - 1} }}dx} $$

$$1 + \cfrac{1}{{1 + \cfrac{2}{{1 + \cfrac{3}{{1 + \cfrac{4}{{1 + ...}}}}}}}} = \cfrac{2}{{1 + \cfrac{1}{{1 + \cfrac{4}{{1 + \cfrac{3}{{... + \cfrac{{2n}}{{1 + \cfrac{{2n - 1}}{{1 + ...}}}}}}}}}}}}$$

$$\int\limits_0^{\frac{{2\pi }}{3}} {{{\ln }^2}\left( {2\cos \frac{x}{2}} \right)dx} $$

$$\operatorname{tg} \left( {\frac{1}{4}\operatorname{arctg} \frac{4}{3}} \right) = \sqrt 5 - 2.$$ $$\operatorname{tg} \left( {\frac{1}{3}\operatorname{arctg} \frac{{\sqrt 5 }}{7}} \right) = \frac{{3\sqrt 3 - 2\sqrt 5 }}{7}$$ $$\operatorname{tg} \left( {\frac{1}{4}\operatorname{arctg} \frac{{\sqrt 5 }}{2}} \right) = \sqrt 6 - \sqrt 5 $$ $$\operatorname{tg} \left( {\frac{1}{4}\operatorname{arctg} \frac{1}{8}} \right) = \frac{{\sqrt 7 }}{7}$$

$$\sum\limits_{k = 0}^{ + \infty } {\dfrac{{{x^{k + 1}}}}{{k{!^s}}}} = x + \dfrac{{{x^2} + \dfrac{{{x^3} + \dfrac{{{x^4} + \dfrac{{{x^5} + ...}}{{{4^s}}}}}{{{3^s}}}}}{{{2^s}}}}}{{{1^s}}},{\text{ }}s > 0$$

\[{\text{Вычислите:}}\] $$\sqrt[4]{{18 - \sqrt[4]{{18 - \sqrt[4]{{18 - ...}}}}}}$$

$$\sqrt[3]{{6 + \sqrt[3]{{6 + \sqrt[3]{{6 + \sqrt[3]{{6 + ...}}}}}}}}=x .$$

$$\prod\limits_{k = 1}^{ + \infty } {\frac{{2k\left( {4k + 3} \right)}}{{\left( {2k + 1} \right)\left( {4k + 1} \right)}}} = \frac{1}{6}{\rm B}\left( {\frac{1}{2},\frac{1}{4}} \right)$$

$$\frac{{{\pi ^2}}}{6} = \frac{3}{2} + \frac{1}{{7 - \frac{{{1^4}}}{{11 - \frac{{{2^4}}}{{19 - \frac{{...}}{{... - \frac{{{n^4}}}{{2{n^2} - 2n + 7 - ...}}}}}}}}}}$$

$$\sum\limits_{k = 0}^{ + \infty } {{{\left( { - 1} \right)}^k}{{\left( {\prod\limits_{i = 1}^k {\frac{{2i - 1}}{{2i}}} } \right)}^2}} = \frac{1}{{2\pi }}{\rm B}\left( {\frac{1}{2},\frac{1}{4}} \right) = \frac{{\sqrt {2\pi } }}{{2\Gamma {{\left( {\frac{3}{4}} \right)}^2}}}.$$

$$\begin{array}{l} \frac{1}{2} + \frac{2}{{1 + 2 \cdot \left( {\frac{1}{4} + \frac{4}{{1 + 4 \cdot \left( {\frac{1}{6} + \frac{6}{{1 + 6 \cdot \left( {...} \right)}}} \right)}}} \right)}} = \frac{{{\pi ^2}}}{8} - 1 + G, \hfill \\ {\text{где }}G{\text{ - постоянная Каталана}}{\text{.}} \hfill \\ \end{array} $$

$$\sum\limits_{k = 0}^{ + \infty } {\frac{{{{\left( { - 1} \right)}^k}}}{{{{\left( {2k + 1} \right)}^3}}}} = \frac{{{\pi ^3}}}{{32}}$$

$$\sum\limits_{k = 1}^{ + \infty } {\frac{1}{{{\pi ^2}{k^2} + 1}}} $$

$$\int\limits_0^{ + \infty } {\frac{{\sin x}}{{{e^{\pi x}} - 1}}dx} = \frac{1}{{{e^2} - 1}}$$

$$\int\limits_0^{ + \infty } {\frac{{{x^{2n}}}}{{{e^{\pi x}} - 1}}dx} = \frac{{\left( {2n} \right)!\zeta \left( {2n + 1} \right)}}{{{\pi ^{2n + 1}}}},{\text{ }}n \in \mathbb{N}.$$

$$\int\limits_0^{ + \infty } {\frac{x}{{{x^{\ln x}}}}dx} $$

\[\begin{array}{l} {\text{Найдите криволинейный интеграл 1 рода }}\int\limits_l {\sqrt {1 - {x^2} - {y^2}} dl} \hfill \\ {\text{по дуге лемнискаты }}{\left( {{x^2} + {y^2}} \right)^2} = {x^2} - {y^2},{\text{ }}x \geqslant 0,{\text{ }}y \geqslant 0. \hfill \\ \end{array} \]

$$\int\limits_0^{ + \infty } {\frac{{\sin \left( {{x^2}} \right)}}{{{e^{{x^2}}}}}dx} = \frac{1}{4}\sqrt {\frac{\pi }{{1 + \sqrt 2 }}} $$
$$\int\limits_0^{ + \infty } {\frac{{\cos \left( {{x^2}} \right)}}{{{e^{{x^2}}}}}dx} = \frac{1}{4}\sqrt {\pi \cdot \left( {1 + \sqrt 2 } \right)} $$

$$\begin{array}{l}\cos \frac{\pi }{{17}} \cdot \cos \frac{{4\pi }}{{17}} = \frac{{1 + \sqrt {17} + \sqrt {34 + 2\sqrt {17} } }}{{16}} \hfill \\ \frac{1}{{\cos \frac{\pi }{{17}}}} + \frac{1}{{\cos \frac{{2\pi }}{{17}}}} - \frac{1}{{\cos \frac{{4\pi }}{{17}}}} + \frac{1}{{\cos \frac{{8\pi }}{{17}}}} = 2\sqrt {17 + 4\sqrt {17} }\end{array} $$

\[\int\limits_0^1 {\frac{{\arctan \left( {{x^k}} \right)}}{x}dx} \]

\[{\text{Найдите }}y \in \mathbb{N}.\] $$\frac{1}{{\sqrt[3]{{{{\cos }^2}\frac{{2\pi }}{9}}}}} + \frac{1}{{\sqrt[3]{{{{\cos }^2}\frac{{4\pi }}{9}}}}} + \frac{1}{{\sqrt[3]{{{{\cos }^2}\frac{{8\pi }}{9}}}}} = \sqrt[3]{{\frac{{y\left( {\sqrt[3]{9} - 1} \right)}}{{\sqrt[3]{9} - 2}}}}$$

\[{\text{Найдите }}x \in \mathbb{N}.\] $$\frac{1}{{\sqrt[3]{{\cos \frac{{2\pi }}{9}}}}} + \frac{1}{{\sqrt[3]{{\cos \frac{{4\pi }}{9}}}}} + \frac{1}{{\sqrt[3]{{\cos \frac{{8\pi }}{9}}}}} = \sqrt[3]{{x\left( {\sqrt[3]{9} - 1} \right)}}$$

$$\mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {\sum\limits_{k = 1}^n {{k^{\frac{1}{3}}}} } \right)}^3} \cdot {{\left( {\sum\limits_{k = 1}^n {{k^{\frac{1}{4}}}} } \right)}^4}}}{{{{\left( {\sum\limits_{k = 1}^n {{k^{\frac{1}{2}}}} } \right)}^2} \cdot {{\left( {\sum\limits_{k = 1}^n {{k^{\frac{1}{5}}}} } \right)}^5}}}$$

$$\begin{array}{l} {\text{Найдите целое число }}x,{\text{ которое удовлетворяет равенству}}{\text{.}} \hfill \\ \sqrt[4]{{3\sqrt 5 - 1 - \sqrt {6\left( {5 - \sqrt 5 } \right)} }} + \sqrt[4]{{3\sqrt 5 - 1 + \sqrt {6\left( {5 - \sqrt 5 } \right)} }} = \sqrt {x + \sqrt {6 + 6\sqrt 5 } } . \hfill \\ \end{array} $$

$$\int\limits_0^{ + \infty } {\frac{1}{{\cosh \sqrt x }}dx} $$

$%\begin{array}{l} {\text{Выпуклый пятиугольник }}ABCDE{\text{ вписан в окружность}}{\text{. }}AB = 1,{\text{ }}BC = 2,{\text{ }}CD = 3,{\text{ }}DE = 4,{\text{ }}AE = \frac{{6\sqrt {14} - 2}}{5}. \hfill \\ {\text{Найдите радиус описанной окружности}}{\text{.}} \hfill \\ \end{array} $%

\[\begin{array}{l} {\text{Пусть }}f\left( x \right) = \frac{{x + 1}}{{x + 2}},{\text{ }}f\left( {f\left( x \right)} \right) = \frac{{{a_2}x + {b_2}}}{{{b_2}x + {c_2}}},{\text{ }}f\left( {f\left( {f\left( x \right)} \right)} \right) = \frac{{{a_3}x + {b_3}}}{{{b_3}x + {c_3}}}{\text{ и т}}{\text{.д}}{\text{.}} \hfill \\ {a_n},{b_n},{c_n} \in \mathbb{N}.{\text{ Докажите}}{\text{, что }}{a_n}{c_n} = b_n^2 + 1. \hfill \\ \end{array} \]

\[{\text{Prove that}}\] $$\sum\limits_{k = 0}^n {\frac{{\left( {2k} \right)!!}}{{\left( {2k + 1} \right)!!}}} = \frac{{2n + 3}}{2} \cdot {\rm B}\left( {\frac{1}{2},n + 2} \right) - 1.$$

$$\int\limits_0^{ + \infty } {\frac{{dx}}{{{{\cosh }^{2n}}x}}} ,{\text{ }}\int\limits_0^{ + \infty } {\frac{{dx}}{{{{\cosh }^{2n + 1}}x}}} $$

$$\int\limits_0^{ + \infty } {\frac{{dx}}{{\sqrt {\cosh x} }}} $$

$$\int\limits_0^{ + \infty } {\frac{{{x^{2n}}}}{{\cosh x}}dx} = \left| {{E_{2n}}} \right| \cdot {\left( {\frac{\pi }{2}} \right)^{2n + 1}},$$ \[n \in \mathbb{N}.\]

$$\int\limits_0^{ + \infty } {\frac{{\left| {\sin x} \right|}}{{{e^x}}}dx} $$

$$\mathop {\lim }\limits_{n \to \infty } \left( {n \cdot \prod\limits_{k = 1}^n {\frac{{1 + {{\left( {2k} \right)}^2}}}{{1 + {{\left( {2k + 1} \right)}^2}}}} } \right) = \tanh \frac{\pi }{2}$$

\[\int\limits_0^{ + \infty } {\frac{x}{{{e^x} + {e^{ - x}}}}dx} \]

\[{\text{Simplify }}\sqrt[n]{{a\sqrt[n]{{a\sqrt[n]{{a\sqrt[n]{{...}}}}}}}},{\text{ }}a > 0.\]

\[{\text{Evaluate }}f\left( x \right) = {x^{{x^{{x^{...}}}}}}{\text{ (infinite power tower) at }}x = {e^{\frac{1}{e}}}.{\text{ Prove that }}f\left( x \right) = + \infty {\text{ if }}x > {e^{\frac{1}{e}}}.\]

\[{\text{Evaluate }}{\left( {\frac{1}{4}} \right)^{{{\left( {\frac{1}{4}} \right)}^{{{\left( {\frac{1}{4}} \right)}^{...}}}}}}{\text{ (infinite power tower)}}{\text{.}}\]

\[\mathop {\lim }\limits_{x \to \frac{1}{2}} \frac{{{{\arcsin }^{\frac{1}{2}}}x - {{\left( {\frac{\pi }{6}} \right)}^{\frac{1}{2}}}}}{{{x^{\frac{1}{2}}} - {{\left( {\frac{1}{2}} \right)}^{\frac{1}{2}}}}}\]

$%\begin{array}{l} {\text{Пусть }}0 < a < 1.{\text{ Найдите уравнение}}{\text{, корнем которого является число }}x = {a^{{a^{{a^{a...}}}}}} \hfill \\ {\text{(бесконечная башня степеней)}}{\text{. (В записи уравнения не должно содержаться }}{a^{{a^{{a^{a...}}}}}}{\text{)}}{\text{.}} \hfill \\ \end{array} $%

\[{a_1} = 2,{\text{ }}{a_2} = 3,{\text{ }}{a_n} = \sqrt {{a_{n - 1}} \cdot {a_{n - 2}}} {\text{ if }}n \geqslant 3.{\text{ Find }}\mathop {\lim }\limits_{n \to \infty } {a_n}.\]