Euler’s Series of Pi

C o n c e p t s

This topic requires familiarity with the following concepts:

Sequences and series
Polynomial expressions
l'Hopital's Rule
Integration and differentiation
Hyperbolic functions
Taylor series
Roots of polynomials

Euler proved that the sum of the infinite reciprocals of the square converge to one-sixth of π squared:

$\dfrac{\pi^2}{6} = \dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...=$ $\sum_{n=1}^{\infty}\dfrac{1}{n^2}$

He found the convergence of other series of the form: $\sum_{n=1}^{\infty}\dfrac{1}{n^{2p}} = K\pi^{2p}$, where n is an integer and K is some real number. Here I will choose to call the series of the form $\sum \dfrac{1}{[(k\pi)^2\pm z^2]^p}$ Euler’s Series of π of order 2p. I will denote its convergence as the following expressions: $S_{2p}(z) = \sum_{n=1}^{\infty} \dfrac{1}{[(n\pi)^2 - z^2]^p}$ and $S_{2pi}(z) = \sum_{n=1}^{\infty} \dfrac{1}{[(n\pi)^2 + z^2]^p}$ (S for the sine function).

After tracing methods Euler used to prove the convergence of his series, I finally found a method to find S_2p(z) and S_2pi(z). It’s rather tedious to find the convergence values for p > 2, but it can be found through diligence if desired. Here, I offer the proof for p = 1 and p = 2. The products of squares are also mentioned.

The following relationships will be proved:

Let $S_{2}(z)=\sum_{n=1}^{\infty}\dfrac{1}{(n\pi)^2-z^2} = \sum_{n=1}^{\infty}\dfrac{\tan (z/2^n)}{z\cdot 2^{n+1}}$. Then $S_{2}(z)=\dfrac{\sin(z)-z\cos(z)}{2z^2\sin(z)}$, z ≠ 0.
Let $S_{2i}(z)=\sum_{n=1}^{\infty}\dfrac{1}{(n\pi)^2+z^2} = \sum_{n=1}^{\infty}\dfrac{\tanh (z/2^n)}{z\cdot 2^{n+1}}$. Then $S_{2i}(z)=\dfrac{z\cosh(z)-\sinh(z)}{2z^2\sinh(z)}$, z ≠ 0.
$S_{4}(z)=\sum_ {n=1}^{\infty}\dfrac{1}{[(n\pi)^2-z^2]^{2}}=$ $\dfrac{z^2 +z\cos(z)\sin(z)-2\sin^{2}(z)}{4z^4\sin^{2}(z)}$, z ≠ 0.
$S_{4i}(z)=\sum_ {n=1}^{\infty}\dfrac{1}{[(n\pi)^2 +z^2]^2}=$ $\dfrac{2\sinh^{2}(z)-z\cosh(z)\sinh(z)-z^2)}{4z^4\sinh^{2}(z)}$, z ≠ 0.

For the case when z = 0, we can evaluate the limit to obtain the same result as Euler as stated in the first equation. Thus, $\underset{z\to 0}{\mathop{\lim}}\,S_2(z)=\underset{z\to 0}{\mathop{\lim}}\,S_2i(z)=\dfrac{1}{6}$, by applying l’Hopital’s rule.

Proof

The Taylor expansion of sine is $\sin x = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}-\dfrac{x^11}{11!}+\text{....}$. Without difficulty, we manipulate to obtain the following critical expression for the proof:

(i) $\dfrac{\sin \sqrt{x}}{\sqrt{x}} = 1-\dfrac{x}{3!}+\dfrac{x^2}{5!}-\dfrac{x^3}{7!}+\dfrac{x^4}{9!}-\dfrac{x^5}{11!}+\text{....}$

The zeros of sin x are x = {0, ±π, ±2π, ±3π, ±4π, ....., ±nπ} for integers n. Thus, the zeros of $\dfrac{\sin \sqrt{x}}{\sqrt{x}}$ are simply the zeros of $\sin \sqrt{x}$, which are x = {π², (2π)², (3π)², (4π)², ....., (nπ)²}.

(Zero is not a root since we get an indeterminate form 0/0.)

Now, we substitute y + z² for x. Therefore:

(ii) $\dfrac{\sin \sqrt{y+z^2}}{\sqrt{y+z^2}} =$ $1-\dfrac{y+z^2}{3!}+\dfrac{(y+z^2)^2}{5!}-$ $\dfrac{(y+z^2)^3}{7!}+$ $\dfrac{(y+z^2)^4}{9!}-\text{....}$

Substitution changes the zeros of (ii) to y = {π² – z², (2π)² – z², (3π)² – z², (4π)² – z², ....., (nπ)² – z²}.

We want to express the series (ii) in the form of a polynomial of the form: $P_{n}(y) = a_0 + a_1 y + a_2 y^2 + a_3 y^3 + a_4 y^4 + \text{....} + a_n y^n$. Euler knew that if P_n(y) is a polynomial of degree n, where n is finite, then the reciprocals of the roots have a sum of $-\dfrac{a_1}{a_0}$. (For a discussion on polynomial roots, see Roots of Polynomials.) He extended this idea to polynomials of infinite terms. Therefore, if we are to find the sum of the reciprocal of the roots in (ii) we must find the coefficient of the y term a₁ and the constant a₀. We will find the coefficient and the constant by manipulating the Taylor series for equation (ii).

The Taylor series for a function f(x) is given by: $f(x)=f(c)+f'(c)(x-c)+\dfrac{f''(c)}{2!}(x-c)^2 + \text{....}$. If we put x = x + c, the series becomes:

(iii) $f(x+c)=$ $f(c)+f'(c)x+\dfrac{f''(c)}{2}x^2 + \dfrac{f'''(c)}{3!}x^3 +$ $\dfrac{f^{(4)}(c)}{4!}x^4 + \text{....}$

Therefore, if we denote P_s(y) as $\dfrac{\sin \sqrt{y}}{\sqrt{y}}$ and make the substitution y = y + z², then we have:

(iv) $\dfrac{\sin \sqrt{y+z^2}}{\sqrt{y+z^2}}=$ $P_{s}(z^2) + \dfrac{P_{s}^{(1)}(z^2)}{1!}y+$ $\dfrac{P_{s}^{(2)}(z^2)}{2!}y^2 + \dfrac{P_{s}^{(3)}(z^2)}{3!}y^3 +$ $\dfrac{P_{s}^{(4)}(z^2)}{4!}y^4 + \text{....}$

For (iv), we have $P_{s}(z^2) = \dfrac{\sin z}{z}$ and $P_{s}^{(1)}(z^2) = \dfrac{d}{dy} \left[y^{-\dfrac{1}{2}}\sin \left( y^{\dfrac{1}{2}} \right) \right]_{y=z^2} = \dfrac{z\cos z-\sin z}{2z^3}$, which are a₀ and a₁, respectively. Hence,

(v) $-\dfrac{a_1}{a_0} = -\dfrac{P_{s}^{(1)}(z^2)}{P_S (z^2)} = -\dfrac{\dfrac{z\cos z-\sin z}{2z^3}}{\dfrac{\sin z}{z}} = \dfrac{\sin z-z\cos z}{2z^2\sin z}$, where z ≠ 0

This proves the first case where the form in the denominator is a difference of two squares. To find the formula for the second case where the form of the denominator is the sum of two squares, use the substitution z = zi, where i is the imaginary unit:

(vi) $\sum\limits_{n=1}^{\infty }{\dfrac{1}{(n\pi)^2}+z^2}=$ $S_{2i}(z)=\dfrac{\sin (zi)-(zi)\cos (zi)}{2(zi)^2\sin (zi)}=$ $-\dfrac{i\sinh (z)-iz\cosh (z)}{2z^2 i\sinh (z)}=\dfrac{z\cosh z-\sinh z}{2z^2\sinh z}$

The last expression was obtained by using the substitutions sin(xi) = isinh(x) and cos(xi) = cosh(x).

Because I used the substitution with i, the notation S_2pi(z) was used for summations of the type $S_{2pi}(z)=\sum_{n=1}^{\infty} \dfrac{1}{[(n\pi )^2 + z^2]^p}$.

Conversion to Product

We observe that (v) can be rewritten as $S_2(z)=\dfrac{\sin z-z\cos z}{2z^2 z}\cdot \dfrac{z}{\sin z}$. This also means that:

(vii) $S_2(\sqrt{z})=$ $\dfrac{\sin \sqrt{z}-\sqrt{z}\cos \sqrt{z}}{2z\sqrt{z}}\cdot \dfrac{\sqrt{z}}{\sin \sqrt{z}}=$ $-\dfrac{\dfrac{d}{dz}\left[ \dfrac{\sin \sqrt{z}}{\sqrt{z}} \right]}{\dfrac{\sin \sqrt{z}}{\sqrt{z}}}$

From (vi), we have:

(viii) $\int{{{S}_{2}}(\sqrt{z})dz=-\ln \dfrac{\sin \sqrt{z}}{\sqrt{z}}}$

Also, $S_{2}(\sqrt{z})=\dfrac{1}{{{\pi }^{2}}-z}+\dfrac{1}{{{(2\pi )}^{2}}-z}+\dfrac{1}{{{(3\pi )}^{2}}-z}+$ $\dfrac{1}{{{(4\pi )}^{2}}-z}+\text{....}$, integration gives:

(ix) $\int{\text{ }{{S}_{2}}(\sqrt{z})\text{ }dz}=$ $-\left[ \ln ({{\pi }^{2}}-z)+\ln (4{{\pi }^{2}}-z)+\ln (9{{\pi }^{2}}-z)+\text{ }..... \right]$

and since (vii) and (viii) are equal we have:

(x) $\left[ \ln \dfrac{\sin \sqrt{z}}{\sqrt{z}} \right]_{z_{1}^{2}}^{z_{2}^{2}}=$ $\left[ \ln [{{\pi }^{2}}-z]+\ln [{{(2\pi )}^{2}}-z]+\ln [{{(3\pi )}^{2}}-z]+\ln [{{(4\pi )}^{2}}-z]+\text{.....} \right]_{z_{1}^{2}}^{z_{2}^{2}}$

Since z cannot be negative, we have to let the lower and upper limits be $z_{1}^{2}$ and $z_{2}^{2}$, respectively. Therefore, using the sum of log identity, we have

(xi) $\dfrac{{{z}_{1}}\sin {{z}_{2}}}{{{z}_{2}}\sin {{z}_{1}}}=$ $\left( \dfrac{{{\pi }^{2}}-z_{2}^{2}}{{{\pi }^{2}}-z_{1}^{2}} \right)\left( \dfrac{{{(2\pi )}^{2}}-z_{2}^{2}}{{{(2\pi )}^{2}}-z_{1}^{2}} \right)$ $\left( \dfrac{{{(3\pi )}^{2}}-z_{2}^{2}}{{{(3\pi )}^{2}}-z_{1}^{2}} \right)\left( \dfrac{{{(4\pi )}^{2}}-z_{2}^{2}}{{{(4\pi )}^{2}}-z_{1}^{2}} \right)\text{.....}=$ $\prod\limits_{n=1}^{\infty }{\dfrac{{{(n\pi )}^{2}}-z_{2}^{2}}{{{(n\pi )}^{2}}-z_{1}^{2}}}$

Making another substitution of z₁ = z₁i and z₂ = z₂i, we obtain:

(xii) $\dfrac{{{z}_{1}}\sinh {{z}_{2}}}{{{z}_{2}}\sinh {{z}_{1}}}=\prod\limits_{n=1}^{\infty }{\dfrac{{{(n\pi )}^{2}}+z_{2}^{2}}{{{(n\pi )}^{2}}+z_{1}^{2}}}$

General Series of Squares

Let’s return to our original function ${{S}_{2}}(z)=\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{(n\pi )}^{2}}-{{z}^{2}}}}$. If we make the substitution z = kπ, where k is any constant, we obtain: ${{S}_{2}}(k\pi )=\dfrac{\sin (k\pi )-(k\pi )\cos (k\pi )}{2{{(k\pi )}^{2}}\sin (k\pi )}=\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{2}}{{\pi }^{2}}-{{k}^{2}}{{\pi }^{2}}}}$ or

(xiii) ${{\pi }^{2}}\cdot {{S}_{2}}(k\pi )=\dfrac{\sin (k\pi )-(k\pi )\cos (k\pi )}{2{{k}^{2}}\sin (k\pi )}=$ $\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{2}}-{{k}^{2}}}}$

Again, a substitution of k = ki would give us

(xiv) ${{\pi }^{2}}\cdot {{S}_{2}}(k\pi )=\dfrac{(k\pi )\cosh (k\pi )-\sinh (k\pi )}{2{{k}^{2}}\sinh (k\pi )}=$ $\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{n}^{2}}+{{k}^{2}}}}$

Order 4p

(xv) $\sum_{n=1}^{\infty } \dfrac{1} {[n^{2}-k^{2}]^{2}} = $ $\dfrac{(k\pi)^{2}+(k\pi)\cos(k\pi )\sin(k\pi)-2\sin^{2}(k\pi)} {4(k\pi)^{4}\sin^{2}(k\pi)}$

(xvi) $\sum_{n=1}^{\infty } \dfrac{1} {[n^{2}+k^{2}]^{2}} = $ $\dfrac{2\sinh^{2}(k\pi) - (k\pi)\cosh(k\pi)\sinh(k\pi) - (k\pi)^{2}} { 4(k\pi )^{4}\sinh^{2}(k\pi) }$

Series from Cosine

Using the cosine series, we come to a simpler summations:

(xvii) ${{C}_{2}}(z)=\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{[(2n-1)\pi ]}^{2}}-{{z}^{2}}}}=\dfrac{\tan ({z}/{2}\;)}{4z}$

(xviii) ${{\pi }^{2}}\cdot {{C}_{2}}(k\pi )=\sum\limits_{n=1}^{\infty }{\dfrac{1}{{{(2n-1)}^{2}}-{{k}^{2}}}}=\dfrac{\pi \tan ({k\pi }/{2}\;)}{4k}$

(xix) $\sum_{n=1}^{\infty} \dfrac{1} {[(n\pi)^{2} -z^{2}]^{2}} = \sum_{n=1}^{\infty} \dfrac{\sec^{2}(z/2^{n}$} {2^{2n+2}\cdot z^{2}} - \dfrac{\tan(z/{2^{n}}\)} {2^{n+2}\cdot z^{3}}\)