Basel problem

 ( Articles Containing Proofs ) 

The Basel problem is a problem in mathematical analysis with relevance to number theory, concerning an infinite sum of inverse squares. It was first posed by Pietro Mengoli in 1650 and solved by Leonhard Euler in 1734, and read on 5 December 1735 in The Saint Petersburg Academy of Sciences. E41 – De summis serierum reciprocarum Since the problem had withstood the attacks of the leading of the day, Euler's solution brought him immediate fame when he was twenty-eight. Euler generalised the problem considerably, and his ideas were taken up more than a century later by Bernhard Riemann in his seminal 1859 paper "On the Number of Primes Less Than a Given Magnitude", in which he defined his zeta function and proved its basic properties. The problem is named after the city of Basel, hometown of Euler as well as of the Bernoulli family who unsuccessfully attacked the problem.

The Basel problem asks for the precise summation of the reciprocals of the square number of the , i.e. the precise sum of the infinite series: $$\backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{1\}\{n^2\}\; =\; \backslash frac\{1\}\{1^2\}\; +\; \backslash frac\{1\}\{2^2\}\; +\; \backslash frac\{1\}\{3^2\}\; +\; \backslash cdots.$$

The sum of the series is approximately equal to 1.644934. The Basel problem asks for the exact sum of this series (in closed form), as well as a proof that this sum is correct. Euler found the exact sum to be $\{\backslash pi^2\}/\{6\}$ and announced this discovery in 1735. His arguments were based on manipulations that were not justified at the time, although he was later proven correct. He produced an accepted proof in 1741.

The solution to this problem can be used to estimate the probability that two large are coprime. Two random integers in the range from 1 to , in the limit as goes to infinity, are relatively prime with a probability that approaches $\{6\}/\{\backslash pi^2\}$, the reciprocal of the solution to the Basel problem.

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Euler's approach Euler's original derivation of the value $\{\backslash pi^2\}/\{6\}$ essentially extended observations about finite and assumed that these same properties hold true for infinite series.

Of course, Euler's original reasoning requires justification (100 years later, Karl Weierstrass proved that Euler's representation of the sine function as an infinite product is valid, by the Weierstrass factorization theorem), but even without justification, by simply obtaining the correct value, he was able to verify it numerically against partial sums of the series. The agreement he observed gave him sufficient confidence to announce his result to the mathematical community.

To follow Euler's argument, recall the Taylor series expansion of the sine function $$\backslash sin\; x\; =\; x\; -\; \backslash frac\{x^3\}\{3!\}\; +\; \backslash frac\{x^5\}\{5!\}\; -\; \backslash frac\{x^7\}\{7!\}\; +\; \backslash cdots$$ Dividing through by gives $$\backslash frac\{\backslash sin\; x\}\{x\}\; =\; 1\; -\; \backslash frac\{x^2\}\{3!\}\; +\; \backslash frac\{x^4\}\{5!\}\; -\; \backslash frac\{x^6\}\{7!\}\; +\; \backslash cdots\; .$$

The Weierstrass factorization theorem shows that the right-hand side is the product of linear factors given by its roots, just as for finite polynomials. Euler assumed this as a heuristic for expanding an infinite degree polynomial in terms of its roots, but in fact it is not always true for general $P(x)$.A priori, since the left-hand-side is a polynomial (of infinite degree) we can write it as a product of its roots as

    & = Ax \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots.
     

\end{align} Then since we know from elementary calculus that $\backslash lim\_\{x\; \backslash rightarrow\; 0\}\; \backslash frac\{\backslash sin(x)\}\{x\}\; =\; 1$, we conclude that the leading constant must satisfy $A\; =\; 1$. This factorization expands the equation into:

                   &= \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots
     

\end{align}

If we formally multiply out this product and collect all the terms (we are allowed to do so because of Newton's identities), we see by induction that the coefficient of is In particular, letting $H\_n^\{(2)\}\; :=\; \backslash sum\_\{k=1\}^n\; k^\{-2\}$ denote a generalized second-order harmonic number, we can easily prove by induction that $x^2\; \backslash prod\_\{k=1\}^\{n\}\; \backslash left(1-\backslash frac\{x^2\}\{\backslash pi^2\}\backslash right)\; =\; -\backslash frac\{H\_n^\{(2)\}\}\{\backslash pi^2\}\; \backslash rightarrow\; -\backslash frac\{\backslash zeta(2)\}\{\backslash pi^2\}$ as $n\; \backslash rightarrow\; \backslash infty$. $$-\backslash left(\backslash frac\{1\}\{\backslash pi^2\}\; +\; \backslash frac\{1\}\{4\backslash pi^2\}\; +\; \backslash frac\{1\}\{9\backslash pi^2\}\; +\; \backslash cdots\; \backslash right)\; =\; -\backslash frac\{1\}\{\backslash pi^2\}\backslash sum\_\{n=1\}^\{\backslash infty\}\backslash frac\{1\}\{n^2\}.$$

But from the original infinite series expansion of , the coefficient of is . These two coefficients must be equal; thus, $$-\backslash frac\{1\}\{6\}\; =\; -\backslash frac\{1\}\{\backslash pi^2\}\backslash sum\_\{n=1\}^\{\backslash infty\}\backslash frac\{1\}\{n^2\}.$$

Multiplying both sides of this equation by −² gives the sum of the reciprocals of the positive square integers. $$\backslash sum\_\{n=1\}^\{\backslash infty\}\backslash frac\{1\}\{n^2\}\; =\; \backslash frac\{\backslash pi^2\}\{6\}.$$

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Generalizations of Euler's method using elementary symmetric polynomials Using formulae obtained from elementary symmetric polynomials,Cf., the formulae for generalized Stirling numbers proved in: this same approach can be used to enumerate formulae for the even-indexed zeta constants which have the following known formula expanded by the Bernoulli numbers: $$\backslash zeta(2n)\; =\; \backslash frac\{(-1)^\{n-1\}\; (2\backslash pi)^\{2n\}\}\{2\; \backslash cdot\; (2n)!\}\; B\_\{2n\}.$$

For example, let the partial product for $\backslash sin(x)$ expanded as above be defined by $\backslash frac\{S\_n(x)\}\{x\}\; =\; \backslash prod\backslash limits\_\{k=1\}^n\; \backslash left(1\; -\; \backslash frac\{x^2\}\{k^2\; \backslash cdot\; \backslash pi^2\}\backslash right)$. Then using known formulas for elementary symmetric polynomials (a.k.a., Newton's formulas expanded in terms of power sum identities), we can see (for example) that

and so on for subsequent coefficients of $x^\{2k\}\; \backslash frac\{S\_n(x)\}\{x\}$. There are other forms of Newton's identities expressing the (finite) power sums $H\_n^\{(2k)\}$ in terms of the elementary symmetric polynomials, $e\_i\; \backslash equiv\; e\_i\backslash left(-\backslash frac\{\backslash pi^2\}\{1^2\},\; -\backslash frac\{\backslash pi^2\}\{2^2\},\; -\backslash frac\{\backslash pi^2\}\{3^2\},\; -\backslash frac\{\backslash pi^2\}\{4^2\},\; \backslash ldots\backslash right),$ but we can go a more direct route to expressing non-recursive formulas for $\backslash zeta(2k)$ using the method of elementary symmetric polynomials. Namely, we have a recurrence relation between the elementary symmetric polynomials and the power sum polynomials given as on this page by $$(-1)^\{k\}k\; e\_k(x\_1,\backslash ldots,x\_n)\; =\; \backslash sum\_\{j=1\}^k\; (-1)^\{k-j-1\}\; p\_j(x\_1,\backslash ldots,x\_n)e\_\{k-j\}(x\_1,\backslash ldots,x\_n),$$

which in our situation equates to the limiting recurrence relation (or generating function convolution, or Cauchy product) expanded as $$\backslash frac\{\backslash pi^\{2k\}\}\{2\}\backslash cdot\; \backslash frac\{(2k)\; \backslash cdot\; (-1)^k\}\{(2k+1)!\}\; =\; -x^\{2k\}\; \backslash frac\{\backslash sin(\backslash pi\; x)\}\{\backslash pi\; x\}\; \backslash times\; \backslash sum\_\{i\; \backslash geq\; 1\}\; \backslash zeta(2i)\; x^i.$$

Then by differentiation and rearrangement of the terms in the previous equation, we obtain that $$\backslash zeta(2k)\; =\; x^\{2k\}\backslash frac\{1\}\{2\}\backslash left(1-\backslash pi\; x\backslash cot(\backslash pi\; x)\backslash right).$$

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Consequences of Euler's proof By the above results, we can conclude that $\backslash zeta(2k)$ is always a rational multiple of $\backslash pi^\{2k\}$. In particular, since $\backslash pi$ and integer powers of it are transcendental, we can conclude at this point that $\backslash zeta(2k)$ is irrational, and more precisely, transcendental for all $k\; \backslash geq\; 1$. By contrast, the properties of the odd-indexed zeta constants, including Apéry's constant $\backslash zeta(3)$, are almost completely unknown.

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The Riemann zeta function The Riemann zeta function is one of the most significant functions in mathematics because of its relationship to the distribution of the . The zeta function is defined for any complex number with real part greater than 1 by the following formula: $$\backslash zeta(s)\; =\; \backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{1\}\{n^s\}.$$

Taking , we see that is equal to the sum of the reciprocals of the squares of all positive integers: $$\backslash zeta(2)\; =\; \backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{1\}\{n^2\}$$

               = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6} \approx 1.644934.
     

Convergence can be proven by the integral test, or by the following inequality: $$\backslash begin\{align\}$$

 \sum_{n=1}^N \frac{1}{n^2} & < 1 + \sum_{n=2}^N \frac{1}{n(n-1)} \\
                            & = 1 + \sum_{n=2}^N \left( \frac{1}{n-1} - \frac{1}{n} \right) \\
                            & = 1 + 1 - \frac{1}{N} \;{\stackrel{N \to \infty}{\longrightarrow}}\; 2.
     

\end{align}

This gives us the upper bound 2, and because the infinite sum contains no negative terms, it must converge to a value strictly between 0 and 2. It can be shown that has a simple expression in terms of the whenever is a positive even integer. With : $$\backslash zeta(2n)\; =\; \backslash frac\{(2\backslash pi)^\{2n\}(-1)^\{n+1\}B\_\{2n\}\}\{2\backslash cdot(2n)!\}.$$

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A proof using Euler's formula and L'Hôpital's rule The normalized sinc function $\backslash text\{sinc\}(x)=\backslash frac\{\backslash sin\; (\backslash pi\; x)\}\{\backslash pi\; x\}$ has a Weierstrass factorization representation as an infinite product: $$\backslash frac\{\backslash sin\; (\backslash pi\; x)\}\{\backslash pi\; x\}\; =\; \backslash prod\_\{n=1\}^\backslash infty\; \backslash left(1-\backslash frac\{x^2\}\{n^2\}\backslash right).$$

The infinite product is analytic, so taking the natural logarithm of both sides and differentiating yields $$\backslash frac\{\backslash pi\; \backslash cos\; (\backslash pi\; x)\}\{\backslash sin\; (\backslash pi\; x)\}-\backslash frac\{1\}\{x\}=-\backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{2x\}\{n^2-x^2\}$$

(by uniform convergence, the interchange of the derivative and infinite series is permissible). After dividing the equation by $2x$ and regrouping one gets $$\backslash frac\{1\}\{2x^2\}-\backslash frac\{\backslash pi\; \backslash cot\; (\backslash pi\; x)\}\{2x\}=\backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{1\}\{n^2-x^2\}.$$

We make a change of variables ($x=-it$): $$-\backslash frac\{1\}\{2t^2\}+\backslash frac\{\backslash pi\; \backslash cot\; (-\backslash pi\; it)\}\{2it\}=\backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{1\}\{n^2+t^2\}.$$

Euler's formula can be used to deduce that $$\backslash frac\{\backslash pi\; \backslash cot\; (-\backslash pi\; i\; t)\}\{2it\}=\backslash frac\{\backslash pi\}\{2it\}\backslash frac\{i\backslash left(e^\{2\backslash pi\; t\}+1\backslash right)\}\{e^\{2\backslash pi\; t\}-1\}=\backslash frac\{\backslash pi\}\{2t\}+\backslash frac\{\backslash pi\}\{t\backslash left(e^\{2\backslash pi\; t\}\; -\; 1\backslash right)\}.$$ or using the corresponding hyperbolic function: $$\backslash frac\{\backslash pi\; \backslash cot\; (-\backslash pi\; i\; t)\}\{2it\}=\backslash frac\{\backslash pi\}\{2t\}\{i\backslash cot\; (\backslash pi\; i\; t)\}=\backslash frac\{\backslash pi\}\{2t\}\backslash coth(\backslash pi\; t).$$

Then $$\backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{1\}\{n^2+t^2\}=\backslash frac\{\backslash pi\; \backslash left(te^\{2\backslash pi\; t\}+t\backslash right)-e^\{2\backslash pi\; t\}+1\}\{2\backslash left(t^2\; e^\{2\backslash pi\; t\}-t^2\backslash right)\}=-\backslash frac\{1\}\{2t^2\}\; +\; \backslash frac\{\backslash pi\}\{2t\}\; \backslash coth(\backslash pi\; t).$$

Now we take the limit as $t$ approaches zero and use L'Hôpital's rule thrice. By Tannery's theorem applied to $\backslash lim\_\{t\backslash to\backslash infty\}\backslash sum\_\{n=1\}^\backslash infty\; 1/(n^2+1/t^2)$, we can interchange the limit and infinite series so that $\backslash lim\_\{t\backslash to\; 0\}\backslash sum\_\{n=1\}^\backslash infty\; 1/(n^2+t^2)=\backslash sum\_\{n=1\}^\backslash infty\; 1/n^2$ and by L'Hôpital's rule

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A proof using Fourier series Use Parseval's identity (applied to the function ) to obtain $$\backslash sum\_\{n=-\backslash infty\}^\backslash infty\; |c\_n|^2\; =\; \backslash frac\{1\}\{2\backslash pi\}\backslash int\_\{-\backslash pi\}^\backslash pi\; x^2\; \backslash ,\; dx,$$ where $$\backslash begin\{align\}$$

 c_n &= \frac{1}{2\pi}\int_{-\pi}^\pi x e^{-inx} \, dx \\[4pt]
     &= \frac{n\pi \cos(n\pi)-\sin(n\pi)}{\pi n^2} i \\[4pt]
     &= \frac{\cos(n\pi)}{n} i \\[4pt]
     &= \frac{(-1)^n}{n} i
     

\end{align}

for , and . Thus,

and $$\backslash sum\_\{n=-\backslash infty\}^\backslash infty\; |c\_n|^2\; =\; 2\backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{1\}\{n^2\}\; =\; \backslash frac\{1\}\{2\backslash pi\}\; \backslash int\_\{-\backslash pi\}^\backslash pi\; x^2\; \backslash ,\; dx.$$

Therefore, $$\backslash sum\_\{n=1\}^\backslash infty\; \backslash frac\{1\}\{n^2\}\; =\; \backslash frac\{1\}\{4\backslash pi\}\backslash int\_\{-\backslash pi\}^\backslash pi\; x^2\; \backslash ,\; dx\; =\; \backslash frac\{\backslash pi^2\}\{6\}$$ as required.

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Another proof using Parseval's identity Given a complete orthonormal basis in the space $L^2\_\{\backslash operatorname\{per\}\}(0,\; 1)$ of L2 periodic functions over $(0,\; 1)$ (i.e., the subspace of square-integrable functions which are also periodic), denoted by $\backslash \{e\_i\backslash \}\_\{i=-\backslash infty\}^\{\backslash infty\}$, Parseval's identity tells us that $$\backslash |x\backslash |^2\; =\; \backslash sum\_\{i=-\backslash infty\}^\{\backslash infty\}\; |\backslash langle\; e\_i,\; x\backslash rangle|^2,$$

where $\backslash |x\backslash |\; :=\; \backslash sqrt\{\backslash langle\; x,x\backslash rangle\}$ is defined in terms of the inner product on this Hilbert space given by $$\backslash langle\; f,\; g\backslash rangle\; =\; \backslash int\_0^1\; f(x)\; \backslash overline\{g(x)\}\; \backslash ,\; dx,\backslash \; f,g\; \backslash in\; L^2\_\{\backslash operatorname\{per\}\}(0,\; 1).$$

We can consider the orthonormal basis on this space defined by $e\_k\; \backslash equiv\; e\_k(\backslash vartheta)\; :=\; \backslash exp(2\backslash pi\backslash imath\; k\; \backslash vartheta)$ such that $\backslash langle\; e\_k,e\_j\backslash rangle\; =\; \backslash int\_0^1\; e^\{2\backslash pi\backslash imath\; (k-j)\; \backslash vartheta\}\; \backslash ,\; d\backslash vartheta\; =\; \backslash delta\_\{k,j\}$. Then if we take $f(\backslash vartheta)\; :=\; \backslash vartheta$, we can compute both that

by elementary calculus and integration by parts, respectively. Finally, by Parseval's identity stated in the form above, we obtain that

    = 2 \sum_{k=1}^{\infty} \frac{1}{(2\pi k)^2}+ \frac{1}{4} \\
    & \implies \frac{\pi^2}{6} = \frac{2 \pi^2}{3} - \frac{\pi^2}{2} = \zeta(2).
     

\end{align}

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Generalizations and recurrence relations Note that by considering higher-order powers of $f\_j(\backslash vartheta)\; :=\; \backslash vartheta^j\; \backslash in\; L^2\_\{\backslash operatorname\{per\}\}(0,\; 1)$ we can use integration by parts to extend this method to enumerating formulas for $\backslash zeta(2j)$ when $j\; >\; 1$. In particular, suppose we let $$I\_\{j,k\}\; :=\; \backslash int\_0^1\; \backslash vartheta^j\; e^\{-2\backslash pi\backslash imath\; k\backslash vartheta\}\; \backslash ,\; d\backslash vartheta,$$

so that integration by parts yields the recurrence relation that

    & = \begin{cases} \frac{1}{j+1}, & k=0; \\[4pt] -\sum\limits_{m=1}^j \frac{j!}{(j+1-m)!} \cdot \frac{1}{(2\pi\imath \cdot k)^m}, & k \neq 0. \end{cases}
     

\end{align}

Then by applying Parseval's identity as we did for the first case above along with the linearity of the inner product yields that

    & = 2 \sum_{m=1}^j \sum_{r=1}^j \frac{j!^2}{(j+1-m)! (j+1-r)!} \frac{(-1)^r}{\imath^{m+r}} \frac{\zeta(m+r)}{(2\pi)^{m+r}} + \frac{1}{(j+1)^2}.
     

\end{align}

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Proof using differentiation under the integral sign It's possible to prove the result using elementary calculus by applying the differentiation under the integral sign technique to an integral due to Freitas: $$I(\backslash alpha)\; =\; \backslash int\_0^\backslash infty\; \backslash ln\backslash left(1+\backslash alpha\; e^\{-x\}+e^\{-2x\}\backslash right)dx.$$

While the primitive function of the integrand cannot be expressed in terms of elementary functions, by differentiating with respect to $\backslash alpha$ we arrive at

$$\backslash frac\{dI\}\{d\backslash alpha\}\; =\; \backslash int\_0^\backslash infty\; \backslash frac\{e^\{-x\}\}\{1+\backslash alpha\; e^\{-x\}+e^\{-2x\}\}dx,$$ which can be integrated by substituting $u=e^\{-x\}$ and decomposing into partial fractions. In the range $-2\backslash leq\backslash alpha\backslash leq\; 2$ the definite integral reduces to

$$\backslash frac\{dI\}\{d\backslash alpha\}\; =\; \backslash frac\{2\}\{\backslash sqrt\{4-\backslash alpha^2\}\}\backslash left\backslash arctan\backslash left(\backslash frac\{\backslash alpha+2\}\{\backslash sqrt\{4-\backslash alpha^2\}\}\backslash right)-\backslash arctan\backslash left(\backslash frac\{\backslash alpha\}\{\backslash sqrt\{4-\backslash alpha^2\}\}\backslash right)\backslash right.$$

The expression can be simplified using the arctangent addition formula and integrated with respect to $\backslash alpha$ by means of trigonometric substitution, resulting in

$$I(\backslash alpha)\; =\; -\backslash frac\{1\}\{2\}\backslash arccos\backslash left(\backslash frac\{\backslash alpha\}\{2\}\backslash right)^2\; +\; c.$$

The integration constant $c$ can be determined by noticing that two distinct values of $I(\backslash alpha)$ are related by

$$I(2)\; =\; 4I(0),$$ because when calculating $I(2)$ we can Factorization $1+2e^\{-x\}+e^\{-2x\}\; =\; (1+e^\{-x\})^2$ and express it in terms of $I(0)$ using the logarithm of a power identity and the substitution $u=x/2$. This makes it possible to determine $c\; =\; \backslash frac\{\backslash pi^2\}\{6\}$, and it follows that

$$I(-2)\; =\; 2\backslash int\_0^\backslash infty\; \backslash ln(1-e^\{-x\})dx\; =\; -\backslash frac\{\backslash pi^2\}\{3\}.$$

This final integral can be evaluated by expanding the natural logarithm into its Mercator series:

$$\backslash int\_0^\backslash infty\; \backslash ln(1-e^\{-x\})dx\; =\; -\; \backslash sum\_\{n=1\}^\backslash infty\; \backslash int\_0^\backslash infty\; \backslash frac\{e^\{-nx\}\}\{n\}dx\; =\; -\backslash sum\_\{n=1\}^\backslash infty\backslash frac\{1\}\{n^2\}.$$

The last two identities imply

$$\backslash sum\_\{n=1\}^\backslash infty\backslash frac\{1\}\{n^2\}\; =\; \backslash frac\{\backslash pi^2\}\{6\}.$$

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Cauchy's proof While most proofs use results from advanced mathematics, such as Fourier analysis, complex analysis, and multivariable calculus, the following does not even require single-variable calculus (until a single limit is taken at the end).

For a proof using the residue theorem, see here.

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History of this proof The proof goes back to Augustin Louis Cauchy (Cours d'Analyse, 1821, Note VIII). In 1954, this proof appeared in the book of Akiva Yaglom and Isaak Yaglom "Nonelementary Problems in an Elementary Exposition". Later, in 1982, it appeared in the journal Eureka, attributed to John Scholes, but Scholes claims he learned the proof from Peter Swinnerton-Dyer, and in any case he maintains the proof was "common knowledge at Cambridge in the late 1960s".; this anecdote is missing from later editions of this book, which replace it with earlier history of the same proof.

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The proof [[File:limit circle FbN.jpeg|thumb|The inequality
$\backslash tfrac\{1\}\{2\}r^2\backslash tan\backslash theta\; >\; \backslash tfrac\{1\}\{2\}r^2\backslash theta\; >\; \backslash tfrac\{1\}\{2\}r^2\backslash sin\backslash theta$
is shown pictorially for any $\backslash theta\; \backslash in\; (0,\; \backslash pi/2)$. The three terms are the areas of the triangle OAC, circle section OAB, and the triangle OAB.

Taking reciprocals and squaring gives
.]] The main idea behind the proof is to bound the partial (finite) sums $$\backslash sum\_\{k=1\}^m\; \backslash frac\{1\}\{k^2\}\; =\; \backslash frac\{1\}\{1^2\}\; +\; \backslash frac\{1\}\{2^2\}\; +\; \backslash cdots\; +\; \backslash frac\{1\}\{m^2\}$$ between two expressions, each of which will tend to as approaches infinity. The two expressions are derived from identities involving the cotangent and cosecant functions. These identities are in turn derived from de Moivre's formula, and we now turn to establishing these identities.

Let be a real number with , and let be a positive odd integer. Then from de Moivre's formula and the definition of the cotangent function, we have $$\backslash begin\{align\}$$

 \frac{\cos (nx) + i \sin (nx)}{\sin^n x} &= \frac{(\cos x + i\sin x)^n}{\sin^n x} \\[4pt]
                                            &= \left(\frac{\cos x + i \sin x}{\sin x}\right)^n \\[4pt]
                                            &= (\cot x + i)^n.
     

\end{align}

From the binomial theorem, we have

Combining the two equations and equating imaginary parts gives the identity $$\backslash frac\{\backslash sin\; (nx)\}\{\backslash sin^n\; x\}\; =\; \backslash Bigg(\; \{n\; \backslash choose\; 1\}\; \backslash cot^\{n\; -\; 1\}\; x\; -\; \{n\; \backslash choose\; 3\}\; \backslash cot^\{n\; -\; 3\}\; x\; \backslash pm\; \backslash cdots\; \backslash Bigg).$$

We take this identity, fix a positive integer , set , and consider for . Then is a multiple of and therefore . So, $$0\; =\; \}\},$$ where $v\_n\; =\; 2n-1\; \backslash mapsto\; \backslash \{1,3,5,7,9,\backslash ldots\backslash \}$.

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See also

• List of sums of reciprocals

• .
• .
• .
• .

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Notes

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External links

• An infinite series of surprises by C. J. Sangwin
• From ζ(2) to Π. The Proof. step-by-step proof
• , English translation with notes of Euler's paper by Lucas Willis and Thomas J. Osler
• Robin Chapman, Evaluating (fourteen proofs)
• target="_blank" rel="nofollow"> Visualization of Euler's factorization of the sine function
• (animated proof based on the above)

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