hcju/proofwikiqa huggingface dataset:
By Riemann Zeta Function as a Multiple Integral,  :$\displaystyle \map \zeta 2 = \int_0^1 \int_0^1 \frac 1 {1 - x y} \rd A$  Let $\tuple {u, v} = \tuple {\dfrac {x + y} 2, \dfrac{y - x} 2}$ so that: :$\tuple {x, y} = \tuple {u - v, u + v}$ Let: :$\size J = \size {\dfrac {\partial \tuple {x, y} } {\partial \tuple {u, v} } } = 2$ Then, by Change of Variables Theorem (Multivariable Calculus): :$\map \zeta 2 = \displaystyle 2 \iint \limits_S \frac {\d u \rd v} {1 - u^2 + v^2}$ where $S$ is the square defined by the coordinates: :$\tuple {0, 0}, \ \tuple {\dfrac 1 2, -\dfrac 1 2}, \ \tuple {1, 0}, \ \tuple {\dfrac 1 2, \dfrac 1 2}$ Exploiting the symmetry of the square and the function over the $u$-axis, we have: :$\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} \! \int_0^u \frac {\d v \rd u} {1 - u^2 + v^2} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac {\d v \rd u} {1 - u^2 + v^2} }$ Factoring $1 - u^2$ gives us: :$\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} \! \int_0^u \frac 1 {1 - u^2} \frac {\d v \rd u} {\frac {v^2} {1 - u^2} + 1} + \int_{\frac 1 2}^1 \! \int_0^{1 - u} \frac 1 {1 - u^2} \frac {\d v \rd u} {\frac {v^2} {1 - u^2} + 1} }$ and letting: :$s = \dfrac v {\sqrt {1 - u^2} }, \rd s = \dfrac 1 {\sqrt {1 - u^2} }$ allows us to make a substitution into each integral, giving: :$\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} \frac 1 {\sqrt {1 - u^2} } \map \arctan {\frac u {\sqrt {1 - u^2} } } \rd u + \int_{\frac 1 2}^1 \frac 1 {\sqrt {1 - u^2} } \map \arctan {\frac {1 - u} {\sqrt {1 - u^2} } } \rd u}$ Consider the right triangle with sides $1$, $x$ and $\sqrt {1 - x^2}$. Applying Pythagoras's Theorem: :$\arcsin x = \arctan \dfrac x {\sqrt {1 - x^2} }$ Let: {{begin-eqn}} {{eqn | l = \theta       | r = \map \arctan {\dfrac{1 - u} {\sqrt{1 - u^2} } }       | c =  }} {{eqn | ll= \leadsto       | l = \tan^2 \theta       | r = \frac {1 - u} {1 + u}       | c =  }} {{eqn | ll= \leadsto       | l = \sec^2 \theta       | r = \frac 2 {1 + u}       | c = Difference of Squares of Secant and Tangent }} {{eqn | ll= \leadsto       | l = \cos 2 \theta       | r = u       | c = Double Angle Formula for Cosine: Corollary 1 }} {{eqn | ll= \leadsto       | l = \theta       | r = \dfrac 1 2 \arccos u       | c = taking arccosines of both sides }} {{eqn | r = \frac \pi 4 - \frac {\arcsin u} 2       | c = Sine of Complement equals Cosine }} {{end-eqn}} This allows us to convert the arctangents from the integrals into arcsines: :$\map \zeta 2 = \displaystyle 4 \paren {\int_0^{\frac 1 2} {\frac {\arcsin u} {\sqrt {1 - u^2} } \rd u} + \int_{\frac 1 2}^1 {\frac 1 {\sqrt {1 - u^2} } \paren {\frac \pi 4 - \frac {\arcsin u} 2} \rd u} }$ Substituting: :$s = \arcsin u$, $\rd s = \dfrac 1 {\sqrt{1 - u^2} }$ into the arcsines, and splitting the second integral: :$\map \zeta 2 = 4 \paren {\dfrac {\pi^2} {72} + \dfrac {\pi^2} {36} } = \dfrac {\pi^2} 6$ {{qed}}
Let: :$\displaystyle P_k = x \prod_{n \mathop = 1}^k \paren {1 - \frac {x^2} {n^2 \pi^2} }$ We note that: {{begin-eqn}} {{eqn | l = P_k - P_{k - 1}       | r = \paren {-\frac {x^3} {k^2 \pi^2} } \prod_{n \mathop = 1}^{k - 1} \paren {1 - \frac {x^2} {n^2 \pi^2} }       | c =  }} {{eqn | r = -\frac {x^3} {k^2 \pi ^2} + \map O {x^5}       | c = Big-O Notation }} {{end-eqn}} '''Examples:''' {{begin-eqn}} {{eqn | l = P_2 - P_1       | r = x \paren {1 - \frac {x^2} {\pi^2} } \paren {\paren {1 - \frac {x^2} {4 \pi^2} } - 1 }       | c =  }} {{eqn | r = -\frac {x^3} {4 \pi ^2} + \frac {x^5} {4 \pi ^4}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = P_3 - P_2       | r = x \paren {1 - \frac {x^2} {\pi^2} } \paren {1 - \frac {x^2} {4 \pi^2} } \paren {\paren {1 - \frac {x^2} {9 \pi^2} } - 1 }       | c =  }} {{eqn | r = -\frac {x^3} {9 \pi ^2} + \frac {5 x^5} {36 \pi ^4} - \frac {x^7} {36 \pi ^4}       | c =  }} {{end-eqn}} By Telescoping Series we find that the coefficient of $x^3$ in $P_k$ is given by: :$(1): \quad \displaystyle -\frac 1 {\pi^2} \sum_{i \mathop = 1}^k \frac 1 {i^2}$ We also have the following two representations of the Sine of x: {{begin-eqn}} {{eqn | l = \sin x       | r = x \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }       | c = Euler Formula for Sine Function }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sin x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots       | c = Power Series Expansion for Sine Function }} {{end-eqn}} So by taking the limit as $k \to \infty$ in $(1)$ and equating with the coefficient of $x^3$ in the Power Series Expansion for Sine Function, we can deduce: :$\displaystyle -\frac 1 {\pi^2} \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = -\frac 1 {3!}$ hence: :$\displaystyle \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = \frac {\pi^2} 6$ {{qed}}
Let $x \in \openint 0 {\dfrac \pi 2}$ and let $n$ be a non-negative integer. {{begin-eqn}} {{eqn | l = \frac {\cos \paren {2 n + 1} x + i \sin \paren {2 n + 1} x} {\sin^{2 n + 1} x}        | r = \frac {\paren {\cos x + i \sin x}^{2 n + 1} } {\sin^{2 n + 1} x}       | c = De Moivre's Formula }} {{eqn | r = \paren {\cot x + i}^{2 n + 1}       | c =  }} {{eqn | r = \sum_{r \mathop = 0}^{2 n + 1} \binom {2 n + 1} r i^r \cot^{2 n + 1 - r} x       | c = Binomial Theorem }} {{eqn | ll= \leadsto       | l = \frac {\sin \paren {2 n + 1} x} {\sin^{2 n + 1} x}       | r = \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} \paren {-1}^r \cot^{2 \paren {n - r} } x       | c = equating imaginary parts }} {{end-eqn}} Let $x_k = \dfrac {k \pi} {2 n + 1}$ for $k = 1, 2, \ldots, n$.  Then: :$\sin \paren {2 n + 1} x_k = 0$ So we have: :$\displaystyle \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} \paren {-1}^r \cot^{2 \paren {n - r} } x_k = 0$ for $k = 1, 2, \ldots, n$. The numbers $x_k$ are all distinct and in the interval $\openint 0 {\dfrac \pi 2}$.  By Shape of Cotangent Function, $\cot x$ is positive and injective in the interval $\openint 0 {\dfrac \pi 2}$. Therefore $\cot^2 x$ is also injective in this interval.  Hence the numbers $c_k = \cot^2 x_k$ are distinct for $k = 1, 2, \ldots, n$.  These numbers are the $n$ distinct roots of the $n$th degree polynomial: :$\displaystyle \map f c := \sum_{r \mathop = 0}^n \binom {2 n + 1} {2 r + 1} (-1)^r c^{n - r}$ By Viète's Formulas, we can calculate the sum of the roots: :$\displaystyle \sum_{k \mathop = 1}^n \cot^2 x_k = \frac {\binom {2 n + 1} 3} {\binom {2 n + 1} 1} = \frac {2 n \paren {2 n - 1} } 6$ From the Difference of Squares of Cosecant and Cotangent $\cot^2 x = \csc^2 x - 1$ we can similarly deduce: :$\displaystyle \sum_{k \mathop = 1}^n \csc^2 x_k = \frac {2 n \paren {2 n - 1} } 6 + n = \frac {2 n \paren {2 n + 2} } 6$ By Shape of Sine Function, $\sin x$ is positive in the interval $\openint 0 {\dfrac \pi 2}$.  So from the Tangent Inequality $\sin x < x < \tan x$ for $x \in \openint 0 {\dfrac \pi 2}$, we can deduce that $\cot^2 x < 1 / x^2 < \csc^2 x$ in the same interval. Summing this inequality from $x_1$ to $x_n$ gives: {{begin-eqn}} {{eqn | l = \frac {2 n \paren {2 n - 1} } 6       | o = <       | r = \sum_{k \mathop = 1}^n \paren {\frac {2 n + 1} {k \pi} }^2       | c =  }} {{eqn | o = <       | r = \frac {2 n \paren {2 n + 2} } 6       | c =  }} {{end-eqn}} or equivalently: {{begin-eqn}} {{eqn | l = \frac {\pi^2} 6 \frac {2 n \paren {2 n - 1} } {\paren {2 n + 1}^2}       | o = <       | r = \sum_{k \mathop = 1}^n \frac 1 {k^2}       | c =  }} {{eqn | o = <       | r = \frac {\pi^2} 6 \frac {2 n \paren {2 n + 2} } {\paren {2 n + 1}^2}       | c =  }} {{end-eqn}} By Combination Theorem for Limits of Functions, the {{LHS}} and {{RHS}} both tend to $\dfrac {\pi^2} 6$ as $n$ tends to infinity.  Therefore by the Squeeze Theorem: :$\map \zeta 2 = \displaystyle \sum_{k \mathop = 1}^\infty \frac 1 {k^2} = \frac {\pi^2} 6$ {{qed}}
From Sum of Reciprocals of Squares of Odd Integers, :$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2n+1}^2} = \frac {\pi^2} 8$ Note that: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}       | r = \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n}^2} + \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2}       | c = }} {{eqn | r = \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} + \frac {\pi^2} 8       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}       | r = \frac {\pi^2} 6 }} {{end-eqn}} {{qed}}
Let the function $f_n$ be defined as: :$(1): \quad f_n \left({x}\right) := \dfrac 1 2 + \cos x + \cos 2 x + \cdots + \cos n x$ By Sum of Cosines of Multiples of Angle: :$(2): \quad f_n \left({x}\right) = \dfrac {\sin \left({\left({2 n + 1}\right) x / 2}\right)} {2 \sin \left({x / 2}\right)}$ Let $E_n$ be defined as: {{begin-eqn}} {{eqn | l = E_n       | r = \int_0^\pi x f_n \left({x}\right) \mathrm d x       | c =  }} {{eqn | n = 3       | r = \frac {\pi^2} 4 + \sum_{k \mathop = 1}^n \left({\frac {\left({-1}\right)^k - 1} {k^2} }\right)       | c = Primitive of $x \cos a x$ and algebra }} {{end-eqn}} The terms for even $k$ on the {{{RHS}} are zero, so: :$\displaystyle (4): \quad \dfrac 1 2 E_{2 n - 1} = \dfrac {\pi^2} 8 - \sum_{k \mathop = 1}^n \dfrac 1 {\left({2 k - 1}\right)^2}$ It remains to be shown that: :$\displaystyle \lim_{n \mathop \to \infty} E_{2 n - 1} = 0$ which will establish: :$\displaystyle (5): \quad \dfrac {\pi^2} 8 = \sum_{k \mathop = 1}^n \dfrac 1 {\left({2 k - 1}\right)^2}$ Using $(2)$, let $g \left({x}\right)$ be the function defined as: :$g \left({x}\right) := \dfrac {\mathrm d} {\mathrm d x} \left({\dfrac {x / 2} {\sin \left({x /2}\right)} }\right)$ Using Integration by Parts, we obtain: :$\displaystyle (6): \quad E_{2 n - 1} = \dfrac 1 {4 n - 1} \left({2 + 2 \int_0^\pi g \left({x}\right) \cos \dfrac {\left({4 n - 1}\right) x} 2 \, \mathrm d x}\right)$ during which we use Limit of $\dfrac {\sin x} x$: Corollary: :$\displaystyle \lim_{x \mathop \to 0} \dfrac {x / 2} {\sin \left({x / 2}\right)} = 1$ We have that $g \left({x}\right)$ is increasing on the interval of integration. Therefore $g \left({x}\right)$ is bounded on the interval $\left[{ 0 \,.\,.\, \pi}\right]$ by $g \left({\pi}\right) = \dfrac \pi 2$. Hence $(5)$ has been established as being true. Now we divide the (strictly) positive integers into even and odd, and use $(5)$ to obtain: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^\infty \frac 1 {k^2}       | r = \sum_{k \mathop = 1}^\infty \frac 1 {\left({2 k}\right)^2} + \sum_{k \mathop = 1}^\infty \frac 1 {\left({2 k - 1}\right)^2}       | c =  }} {{eqn | r = \frac 1 4 \sum_{k \mathop = 1}^\infty \frac 1 {k^2} + \frac {\pi^2} 8       | c =  }} {{eqn | ll= \leadsto       | l = \frac 3 4 \sum_{k \mathop = 1}^\infty \frac 1 {k^2}       | r = \frac {\pi^2} 8       | c =  }} {{eqn | ll= \leadsto       | l = \sum_{k \mathop = 1}^\infty \frac 1 {k^2}       | r = \frac 4 3 \frac {\pi^2} 8       | c =  }} {{eqn | r = \frac {\pi^2} 6       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \zeta \left({2}\right)       | r = \left({-1}\right)^2 \dfrac {B_2 2^1 \pi^2} {2!}       | c = Riemann Zeta Function at Even Integers }} {{eqn | r = \left({-1}\right)^2 \left({\dfrac 1 6}\right) \dfrac {2^1 \pi^2} {2!}       | c = {{Defof|Sequence of Bernoulli Numbers}} }} {{eqn | r = \left({\dfrac 1 6}\right) \left({\dfrac 2 2}\right) \pi^2       | c = {{Defof|Factorial}} }} {{eqn | r = \dfrac {\pi^2} 6       | c = simplifying }} {{end-eqn}} {{qed}}
By Fourier Series of $x^2$, for $x \in \openint {-\pi} \pi$: :$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos n x}$ Letting $x \to \pi$ from the left: {{begin-eqn}} {{eqn | l = \pi^2       | r = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \frac 4 {n^2} \cos \pi x} }} {{eqn | l = \pi^2       | r = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\paren {-1}^n \paren {-1}^n \frac 4 {n^2} }       | c = Cosine of Multiple of $\pi$ }} {{eqn | l = \pi^2       | r = \frac {\pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} }} {{eqn | ll= \leadsto       | l = \frac {2 \pi^2} 3        | r = 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}       | r = \frac {\pi^2} 6 }} {{end-eqn}} {{qed}}
By Fourier Series of Identity Function over $-\pi$ to $\pi$, for $x \in \openint {-\pi} \pi$: :$\displaystyle x \sim 2 \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1} } n \map \sin {n x}$ Hence: {{begin-eqn}} {{eqn | l = \frac 1 \pi \int_{-\pi}^\pi x^2 \rd x        | r = \sum_{n \mathop = 1}^\infty \paren {\frac{2 \paren {-1}^{n + 1} } n}^2       | c = Parseval's Theorem }} {{eqn | ll= \leadsto       | l = \frac 2 \pi \int_0^\pi x^2 \rd x        | r = \sum_{n \mathop = 1}^\infty \frac 4 {n^2}       | c = Definite Integral of Even Function }} {{eqn | ll=\leadsto       | l = \frac {2 \pi^2} 3        | r = \sum_{n \mathop = 1}^\infty \frac 4 {n^2}  }} {{eqn | ll=\leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}        | r = \frac {\pi^2} 6  }}  {{end-eqn}} {{qed}}
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as: :$\map f x = \begin {cases} \paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end {cases}$ From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as: :$\map f x \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n x} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n x}$ Setting $x = 0$: {{begin-eqn}} {{eqn | l = \map f 0       | r = \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos 0} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin 0}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {0 - \pi}^2       | r = \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos 0} {n^2} }       | c = Sine of Zero is Zero }} {{eqn | ll= \leadsto       | l = \pi^2       | r = \frac {2 \pi^2} 3 + 2 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}       | c = Cosine of Zero is One }} {{eqn | ll= \leadsto       | l = \frac {\pi^2} 2 - \frac {\pi^2} 3       | r = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\pi^2} 6       | r = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}       | c =  }} {{end-eqn}} {{qed}}
From Mittag-Leffler Expansion for Hyperbolic Cotangent Function, we have:  :$\displaystyle \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z} = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$ We can write:  {{begin-eqn}} {{eqn	| l = \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z} 	| r = \frac 1 {2 z} \paren {\frac {\pi \paren {e^{\pi z} + e^{-\pi z} } } {e^{\pi z} - e^{-\pi z} } - \frac 1 z} 	| c = {{Defof|Hyperbolic Cotangent}} }} {{eqn	| r = \frac 1 {2 z} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {z \paren {e^{2 \pi z} - 1} } } }} {{end-eqn}} We therefore have:  :$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }$ We have at $z = 0$:  {{begin-eqn}} {{eqn	| l = \pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1 	| r = \pi \times 0 \times \paren {e^0 + 1} - e^0 + 1 }} {{eqn	| r = 1 - e^0 }} {{eqn	| r = 0 	| c = Exponential of Zero }} {{end-eqn}} and:  :$2 z^2 \paren {e^{2 \pi z} + 1} = 0$ So by L'Hopital's Rule: {{begin-eqn}} {{eqn	| l = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } } 	| r = \lim_{z \mathop \to 0} \paren {\frac {\pi \paren {e^{2 \pi z} + 1} + 2 \pi^2 z e^{2 \pi z} - 2 \pi e^{2 \pi z} } {4 z \paren {e^{2 \pi z} - 1} + 4 \pi z^2 e^{2 \pi z} } } 	| c = Product Rule, Derivative of Exponential Function, Derivative of Power }} {{eqn	| r = \lim_{z \mathop \to 0} \paren {\frac {2 \pi^2 e^{2 \pi z} + 2 \pi^2 e^{2 \pi z} + 4 \pi^3 z e^{2 \pi z} - 4 \pi^2 e^{2 \pi z} } {4 \paren {e^{2 \pi z} - 1} + 8 \pi z e^{2 \pi z} + 8 \pi z e^{2 \pi z} + 8 \pi^2 z^2 e^{2 \pi z} } } 	| c = L'Hopital's Rule }} {{eqn	| r = 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {z e^{2 \pi z} } {-4 + e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} } } }} {{eqn	| r = 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {e^{2 \pi z} + 2 \pi z e^{2 \pi z} } {2 \pi e^{2 \pi z} \paren {8 \pi^2 z^2 + 16 \pi z + 4} + e^{2 \pi z} \paren {16 \pi z + 16 \pi} } } 	| c = L'Hopital's Rule }} {{eqn	| r = 4 \pi^3 \lim_{z \mathop \to 0} \paren {\frac {1 + 2 \pi z} {16 \pi^3 z^2 + 32 \pi^2 z + 8 \pi + 16 \pi z + 16 \pi} } }} {{eqn	| r = \frac {4 \pi^3} {24 \pi} }} {{eqn	| r = \frac {\pi^2} 6 }} {{end-eqn}} giving:  :$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6$ {{qed}}
From the Euler Formula for Sine Function, we have: :$\displaystyle \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$ Therefore, if we divide by x, we get :$\displaystyle \dfrac {\sin x} {x} = \dfrac {x} {x} \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} } = \paren {1 - \frac {x^2} {1^2 \pi^2} } x \paren {1 - \frac {x^2} {2^2 \pi^2} } x \paren {1 - \frac {x^2} {3^2 \pi^2} } x \cdots$ :$\displaystyle  $ From the Power Series Expansion for Sine Function, we have: :$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1} } {\left({2 n + 1}\right)!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \frac {x^7} {7!} + \cdots$ Therefore, if we divide by x, we get :$\displaystyle \dfrac {\sin x} {x}  = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n} } {\left({2 n + 1}\right)!} = 1 - \frac {x^2} {3!} + \frac {x^4} {5!} - \frac {x^6} {7!} + \cdots$ :$\displaystyle  $ Equating the infinite product with the infinite sum, we have: :$\displaystyle \dfrac {\sin x} {x} = \dfrac {x} {x} \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} } = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n} } {\left({2 n + 1}\right)!}$ :$\displaystyle  $ :$\displaystyle \dfrac {\sin x} {x} = \paren {1 - \frac {x^2} {1 \pi^2} } x \paren {1 - \frac {x^2} {4 \pi^2} } x \paren {1 - \frac {x^2} {9 \pi^2} } x \cdots = 1 - \frac {x^2} {3!} + \frac {x^4} {5!} - \frac {x^6} {7!} + \cdots$ :$\displaystyle  $ From this, we can see the following relationships: :$\displaystyle - \frac {x^2} {3!} = - \frac {x^2} { \pi^2}  \paren {1 + \frac {1} {4} + \frac {1} {9} + \frac {1} {16} \cdots} ($"Choose 1" - used to calculate $ \map \zeta 2)$ :$\displaystyle  $ :$\displaystyle  + \frac {x^4} {5!} =  +\frac {x^4} { \pi^4}   \paren {\paren {1} \paren {\frac {1} {4}} + \paren {1} \paren {\frac {1} {9}} + \paren {1} \paren {\frac {1} {16}} + \cdots  + \paren {\frac {1} {4}} \paren {\frac {1} {9}} + \paren {\frac {1} {4}} \paren {\frac {1} {16}} + \cdots  +  \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots} ($"Choose 2" - used to calculate $ \map \zeta 4)$ :$\displaystyle  $ :$\displaystyle - \frac {x^6} {7!} = - \frac {x^6} { \pi^6}   \paren {\paren {1} \paren {\frac {1} {4}} \paren {\frac {1} {9}} + \paren {1} \paren {\frac {1} {4}} \paren {\frac {1} {16}} + \cdots + \paren {1} \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots +  \paren {\frac {1} {4}} \paren {\frac {1} {9}} \paren {\frac {1} {16}} + \cdots} ($"Choose 3" - used to calculate $\map \zeta 6)$ :$\displaystyle  $ The equations above can be used to calculate the zeta function at even integer arguments.  Notice that "Choose $1$" is the summation of each term chosen once. "Choose $2$" is the summation of the product of every non-repeating combination of two terms. "Choose $3$" is the summation of the product of every non-repeating combination of three terms. :$\displaystyle  $ For example, from "Choose $1$" above: :$\displaystyle  \map \zeta 2  = \dfrac { \pi^2} {3!}$ {{qed}}
Proof by induction: === Basis for the Induction === $n = 1$ holds trivially. Just to make sure, we try $n = 2$: :$1 + 2 + 1 = 4$ Likewise $n^2 = 2^2 = 4$. So shown for basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :$1 + 2 + \cdots + k + \paren {k - 1} + \cdots + 1 = k^2$ Now we need to show true for $n = k + 1$: :$1 + 2 + \cdots + \paren {k + 1} + k + \paren {k - 1} + \cdots + 1 = \paren {k + 1}^2$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | o =        | r = 1 + 2 + \cdots + \paren {k + 1} + k + \paren {k - 1} + \cdots + 1       | c =  }} {{eqn | r = \paren {1 + 2 + \cdots + k + \paren {k - 1} + \cdots + 1} + k + \paren {k + 1}       | c =  }} {{eqn | r = k^2 + k + \paren {k + 1}       | c = from induction hypothesis }} {{eqn | r = k^2 + 2k + 1       | c =  }} {{eqn | r = \paren {k + 1}^2       | c =  }} {{end-eqn}} The result follows by induction. {{qed}}
{{begin-eqn}} {{eqn | o =       | r = 1 + 2 + \cdots + \paren {n - 1} + n + \paren {n - 1} + \cdots + 1       | c =  }} {{eqn | r = 1 + 2 + \cdots + \paren {n - 1} + \paren {n - 1} + \cdots + 1 + n       | c =  }} {{eqn | r = 2 \paren {1 + 2 + \cdots + \paren {n - 1} } + n       | c =  }} {{eqn | r = 2 \paren {\frac {\paren {n - 1} n} 2} + n       | c = Closed Form for Triangular Numbers }} {{eqn | r = \paren {n - 1} n + n       | c =  }} {{eqn | r = n^2 - n + n       | c =  }} {{eqn | r = n^2       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = 1 + 2 + \cdots + \paren {n - 1} + n + \paren {n - 1} + \cdots + 1       | c =  }} {{eqn | r = \paren {1 + 2 + \cdots + \paren {n - 1} } + \paren {1 + 2 + \cdots + \paren {n - 1} + n}       | c =  }} {{eqn | r = \frac {\paren {n - 1} n} 2 + \frac {n \paren {n + 1} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {n^2 - n + n^2 + n} 2       | c =  }} {{eqn | r = n^2       | c =  }} {{end-eqn}} {{qed}}
Let $T_n = 1 + 2 + \cdots + n + \left({n - 1}\right) + \cdots + 1$. We have $T_1 = 1$ and {{begin-eqn}} {{eqn | l = T_n - T_{n-1}       | r = \left({1 + 2 + \cdots + n + \left({n - 1}\right) + \cdots + 1 }\right)       | c = Definition of $T_n$ }} {{eqn | o =        | ro= -        | r = \left({1 + 2 + \cdots + \left({n - 1}\right) + \left({n - 2}\right) + \cdots + 1}\right)       | c =  }} {{eqn | r = \left({\left({1 + 2 + \cdots + n}\right) - \left({1 + 2 + \cdots + \left({n - 1}\right)}\right)}\right)       | c = Integer Addition is Associative }} {{eqn | o =        | ro= +        | r = \left({\left({\left({n - 1}\right) + \left({n - 2}\right) + \cdots + 1}\right) - \left({\left({n - 2}\right) + \left({n - 3}\right) + \cdots + 1}\right)}\right)       | c = Integer Addition is Commutative }} {{eqn | r = n + \left({n - 1}\right)       | c = simplifying }} {{eqn | r = 2 n - 1       | c =  }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l = T_n       | r = \left({T_n - T_{n-1} }\right) + \left({T_{n-1} - T_{n-2} }\right) + \cdots + \left({T_2 - T_1}\right) + T_1 }} {{eqn | r = \left({2 n - 1}\right) + \left({2 \left({n - 1}\right) - 1}\right) + \cdots + \left({2 \times 2 - 1}\right) + 1 }} {{eqn | r = \sum_{k \mathop = 1}^n 2 k - 1 }} {{eqn | r = n^2        | c = Odd Number Theorem }} {{end-eqn}} {{qed}}
Let $S$ be a Skolem sequence of order $n$. Let $a_i$ and $b_i$ be the positions of the first and second occurrences respectively of the integer $i$ in $S$, where $1 \le i \le n$. We can thus conclude that: :$b_i - a_i = i$ for each $i$ from 1 to $n$. Summing both sides of this equation we obtain: :$\displaystyle \sum_i^n b_i - \sum_i^n a_i = \sum_i^n i = \frac {n \paren {n + 1} } 2$ Now the $a_i$ and the $b_i$ represent the positions in the sequence from $1$ to $2 n$. Hence: :$\displaystyle \sum_i^n b_i + \sum_i^n a_i = \frac {2 n \paren {2 n + 1} } 2 = n \paren {2 n + 1}$ Summing the previous two equations we obtain the identity: :$\displaystyle \sum_i^n b_i = \frac {n \paren {5 n + 3} } 4$ The {{LHS}} of this last equality is a sum of positions and thus must be an integer. We conclude that the right hand side must also be an integer. This occurs exactly when: $n \equiv 0, 1 \pmod 4$ {{qed}}