Revision as of 16:03, 16 January 2015

Problem

Show, by using the Fourier series, that $$\sum_{k=1}^\infty \frac{1}{k^n} = - \frac{(2 \pi i)^n}{2(n!)} B_n,\ n \in 2 \mathbb{Z}_{>0}$$ where \(B_n\) are the Bernoulli numbers defined as $$\frac{z}{e^z - 1} = \sum_{n=0}^\infty B_n \frac{z^n}{n!}$$

Hint: consider \( B_n(x) := \sum_{k \in \mathbb{Z} \setminus \{0\}} k^{-n} e^{ikx},\ n \in \mathbb{Z}_{\geq 2},\ x \in \mathbb{R} \)

Solution Sketch

The solution below is extremely comprehensive, if not expletive, because I felt that the Musterlösung of the MMP Series that treated the problem was lacking some fairly important steps/guidelines. Therefore find a TL;DR here:

1. See that \(B_n(0) = 2 \sum_{k=1}^\infty k^{-n}\). We now want to find the \(B_n(0)\) for \(n \in 2\mathbb{Z}_{>0}\)
2. We find a polynomial with Fourier coefficients \(c_0=0\) and \(c_k = \frac{1}{k^2}\).
   On \([0,2\pi]\), this polynomial will thus be equal to \(B_2(x)\)
3. \(B'_n(x) = iB_{n-1}(x)\), therefore all \(B_n(x)\) will be polynomials, too. We find a recursion relation for \(B_n(x)\)
4. Using the \(2\pi\)-periodicity of \(B_n(x)\) (for all \(n \geq 2\)), we find a recursion relation for the \(B_n(0)\)
5. We prove the postulated identity by relating the definition for the \(B_n\) given in the problem to the recursion for the \(B_n(0)\) found above.

Solution

As suggested in the problem, we will have a look at the convergent fourier series $$B_n(x) := \sum_{k \in \mathbb{Z} \setminus \{0\}} k^{-n} e^{ikx},\ n \in \mathbb{Z}_{\geq 2},\ x \in \mathbb{R}$$ with Fourier coefficients $$c_k = \begin{cases} 0 & \text{for } k = 0\\ \frac{1}{k^n} & \text{for } k \neq 0 \end{cases}$$

It is obvious that \( B_n(x) \) is \(2\pi\)-periodic and, by differentiating every term separately, that $$B'_n(x) = iB_{n-1}(x)$$

Also, for \(n \in 2\mathbb{Z}_{>0}\), we see that $$B_n(0) = \sum_{k \in \mathbb{Z} \setminus \{0\}} k^{-n} e^{ik\,0} = \sum_{k=-\infty}^{-1} k^{-n} + \sum_{k=1}^\infty k^{-n} = 2 \sum_{k=1}^\infty k^{-n} \tag{$\ddagger$}$$ where the last step follows from the fact that \(n\) is even.

We will first show that \( B_2(x) = f(x) := \pi^2/3 - \pi x + x^2/2 \) for \( x \in [0, 2\pi]\).

Recall the definition of the Fourier coefficients for \(2\pi\)-periodic functions in \(C^\infty(\mathbb{R})\): $$c_k = \frac{1}{2\pi} \int_0^{2\pi} f(x) \: e^{-ikx} \: dx$$

We can easily verify the following identities: $$\begin{align} \frac{1}{2\pi} \int_0^{2\pi} e^{-ikx} \: dx &\ =\ \begin{cases} 1 & \text{for } k = 0\\ 0 & \text{for } k \neq 0 \end{cases} \\ \frac{1}{2\pi} \int_0^{2\pi} x \: e^{-ikx} \: dx &\ =\ \begin{cases} \pi & \text{for } k = 0\\ \frac{i}{k} & \text{for } k \neq 0 \end{cases} \\ \frac{1}{2\pi} \int_0^{2\pi} x^2 \: e^{-ikx} \: dx &\ =\ \begin{cases} \frac{4}{3} \pi^2 & \text{for } k = 0\\ \frac{2(1+ik\pi)}{k^2} & \text{for } k \neq 0 \end{cases} \end{align}$$

Calculating the Fourier coefficients for \(f(x)\) then yields $$\begin{align} c_k &= \frac{1}{2\pi} \int_0^{2\pi} f(x) \: e^{-ikx} \: dx\\ &= \frac{1}{2\pi} \left( \frac{\pi^2}{3} \int_0^{2\pi} e^{-ikx} \: dx - \pi \int_0^{2\pi} x \: e^{-ikx} \: dx + \frac{1}{2} \int_0^{2\pi} x^2 \: e^{-ikx} \right) \\ &= \begin{cases} 0 & \text{for } k = 0\\ \frac{1}{k^2} & \text{for } k \neq 0 \end{cases} \end{align}$$ which proves that \(B_2(x) = f(x)\) for \(x \in [0, 2\pi]\).

Thus, we have shown that the \(B_n|_{[0, 2\pi]}\) are polynomials, which are recursively defined by \(B'_n = iB_{n-1}\) and \(\int_0^{2\pi} B_n(x) dx = -i(B_{n+1}(2\pi) - B_{n+1}(0)) = 0\). We also add $$B_1(x) = -iB'_2(x) = i\pi - ix \qquad B_0(x) = -iB'_1(x) = -1$$

By induction over \(n\), we will now prove that $$B_n(x) = \sum_{k=0}^n B_{n-k}(0) \frac{(ix)^k}{k!}$$ Base case for \(n=0\): $$B_0(0) \frac{(ix)^0}{0!} = B_0(0) = -1 = B_0(x)$$ Inductive step (\(n\!-\!1 \rightarrow n\)): $$\begin{align} B_n(x) &= \int_0^x iB_{n-1}(\tilde x)\ d\tilde x + B_n(0) \\ &= \int_0^x i\left(\sum_{k=0}^{n-1} B_{(n-1)-k}(0) \frac{(i\tilde x)^k}{k!} \right) d\tilde x + B_n(0) \\ &= \left( \sum_{k=0}^{n-1} B_{(n-1)-k}(0) \frac{i^{k+1}}{k!} \int_0^x \tilde x^k\ d\tilde x \right) + B_n(0) \\ &= \left( \sum_{k=0}^{n-1} B_{(n-1)-k}(0) \frac{i^{k+1}}{k!} \frac{x^{k+1}}{k+1} \right) + B_n(0) \\ &= \left( \sum_{k=0}^{n-1} B_{n-(k+1)}(0) \frac{(ix)^{k+1}}{(k+1)!} \right) + B_n(0) \\ &= \left( \sum_{k=1}^n B_{n-k}(0) \frac{(ix)^k}{k!} \right) + B_n(0) \\ &= \sum_{k=0}^n B_{n-k}(0) \frac{(ix)^k}{k!} \end{align}$$

As \(B_n\) is \(2\pi\)-periodic for all \(n \geq 2\), the above identity leads us to the relation $$0 = B_n(2\pi) - B_n(0) = \sum_{k=1}^n B_{n-k}(0) \frac{(2\pi i)^k}{k!} \tag{*}$$ for all \(n \geq 2\), which determines all \(B_n(0)\) from \(B_0(0) = -1\)

The definition of the \(B_n\) in the problem (the Bernoulli numbers, not the functions!) can be transformed to: $$\begin{align} & \frac{z}{e^z - 1} = \sum_{n=0}^\infty B_n \frac{z^n}{n!} \\ \Leftrightarrow\ & z = \left( \sum_{n=0}^\infty B_n \frac{z^n}{n!} \right) \left( \sum_{k=1}^\infty \frac{z^k}{k!} \right) \\ \Leftrightarrow\ & z = \left( B_0 + B_1z + \frac{B_2}{2}z^2 + \frac{B_3}{3!}z^3 + ... \right) \left( z + \frac{z^2}{2} + \frac{z^3}{3!} + \frac{z^4}{4!} + ... \right) \\ \Leftrightarrow\ & z = B_0z + \left( \frac{B_0}{2!} + B_1 \right)z^2 + \left( \frac{B_0}{3!} + \frac{B_1}{2!} + \frac{B_2}{2!} \right)z^3 + \left( \frac{B_0}{4!} + \frac{B_1}{3!} + \frac{B_2}{2!2!} + \frac{B_3}{3!} \right)z^4 \: +\ ... \end{align}$$ which, through equating coefficients yields \(B_0 = 1\) and for \(n \geq 2\) $$\sum_{k=1}^n \frac{B_{n-k}}{(n-k)!k!} = 0 \tag{$\dagger$}$$

The following proof by induction doesn't appear in the aforementioned Musterlösung, where the final step simply falls out of thin air. It is not unlikely, however, that I simply don't understand what they do. If this is the case, feel free to simplify the following mess of equations. --Nik

We now now prove by induction that for \(n \geq 0\) $$B_n(0) = - \frac{(2\pi i)^n B_n}{n!}$$

Base case for \(n=0\): $$B_0(0) = -1 = -B_0$$ Inductive step (\(n\!-\!1 \rightarrow n\)): $$\begin{align} \sum_{k=1}^{n+1} B_{n+1-k}(0) \frac{(2\pi i)^k}{k!} \ &\overset{(*)}{=} \ 0 \\ \Leftrightarrow \ B_n(0)\: 2\pi i \ &= \: - \sum_{k=2}^{n+1} B_{n+1-k}(0) \frac{(2\pi i)^k}{k!} \\ \overset{\text{Hyp.}}{\Leftrightarrow} \ B_n(0)\: 2\pi i \ &= \: - \sum_{k=2}^{n+1} \left( - \frac{(2\pi i)^{n+1-k} B_{n+1-k}}{(n+1-k)!} \right) \frac{(2\pi i)^k}{k!} \\ \Leftrightarrow \ B_n(0)\: 2\pi i \ &= \ (2\pi i)^{n+1} \sum_{k=2}^{n+1} \frac{B_{n+1-k}}{(n+1-k)!\:k!} \\ \overset{(\dagger)}{\Leftrightarrow} \ B_n(0)\: 2\pi i \ &= \ (2\pi i)^{n+1} \left(- \frac{B_n}{n!} \right) \\ B_n(0) \ &= \: - \frac{(2\pi i)^n B_n}{n!} \end{align}$$

This, together with equation \((\ddagger)\), proves the identity in the problem.

\(\square\)

Alternate Solution

This solution has not yet been proof-read, but was proposed by Alessio (aka Milfhunter, veeery classy) as their TA solved the same problem in their tutor class.

Let $$\zeta(s) = \sum\limits_{n=1}^\infty \frac{1}{n^s} $$ for $$s \in \mathbb{C}, \Re (s) > 1$$

Consider $$ E(z) := \frac{z}{e^z -1} := \sum\limits_{n=0}^\infty B_n \frac{z^n}{n!} \quad\quad\quad z \in \mathbb{R} $$ $$\lim\limits_{z \to 0}{E(z)} = \lim\limits_{z \to 0}{\frac{1}{e^z}} = 1 = \lim\limits_{z \to 0}{\sum\limits_{n=0}^\infty B_n \frac{z^n}{n!}} $$ Where in the first equality we used l'Hospital's rule and with $$ \lim\limits_{z \to 0}{E(z)} = E(\lim\limits_{z \to 0}{z}) = \sum\limits_{n=0}^\infty \lim\limits_{z \to 0}{B_n \frac{z^n}{n!}} =B_0 $$ Where we can put the limit inside because E is continuous and bounded in a neighborhood of z_0 = 0. $$ \rightarrow B_0 = 1 $$ Now we repeat the same thing for E' \begin{align*} \lim\limits_{z \to 0}{E'(z)} & = \lim\limits_{z \to 0}{\frac{e^z -1 -z e^z}{e^{2z} -2e^z+1}} \\ & = \lim\limits_{z \to 0}{\frac{e^z -z e^z -e^z}{2e^{2z} -2e^z}} \\ & = \lim\limits_{z \to 0}{\frac{-z e^z - e^z}{4e^{2z}-2e^z}} \\ & = -\frac{1}{2} \end{align*} Here we again used l'Hospital's rule for the second and third equality. For the same reasons as above we get $$ \lim\limits_{z \to 0}{E'(z)} = B_1 $$ and therefore $$\rightarrow b_1 = - \frac{1}{2}$$

$$\Rightarrow E(z) = 1 - \frac{1}{2} z + \mathcal{O}(z^2)$$

Now define \(f(z) = \frac{z}{2} \coth (\frac{z}{2})\) on \(]-\frac{1}{2}, \frac{1}{2}]\) \begin{align*} f(z) & = \frac{z}{2} \coth (\frac{z}{2}) \\ & = \frac{z e^{\frac{z}{2}} + z e^{-\frac{z}{2}}}{2e^{\frac{z}{2}} - 2 e^{-\frac{z}{2}}} \\ & = \frac{z e^{\frac{z}{2}} + z e^{-\frac{z}{2}}}{2e^{\frac{z}{2}}(e^z-1)} \\ & = \frac{z}{2} \frac{e^z+1}{e^z-1} \\ & = \frac{z}{2} \left( 1 + \frac{2}{e^z-1} \right) \\ & = \frac{z}{2} + \frac{z}{e^z-1} \\ & = \frac{z}{2} + E(z) \end{align*}

Since z and coth are odd functions, f(z) is an even function. From the part above we know that $$E(z) = 1 - \frac{1}{2} z + \mathcal{O}(z^2)$$ Since the addition of z/2 kills the uneven term and f(z) has to be even, the only remaining coefficients are even: $$ f(z) = \frac{z}{2} + E(z) = \sum\limits_{n=0}^\infty B_{2n} \frac{z^{2n}}{(2n)!} $$

Now define \(g(x) = \cosh (z x)\) on \(]-\frac{1}{2}, \frac{1}{2}]\) with \(z \in \mathbb{R} \setminus \{0\}\), continued periodically on \(\mathbb{R}\) Calculating the Fourier coefficients of g:

$$\alpha_0 = \int\limits_{-\frac{1}{2}} ^{\frac{1}{2}} \cosh(zx)dx = \frac{2}{z}\sinh (\frac{z}{2})$$

\begin{align*} \alpha_n & = \int\limits_{-\frac{1}{2}} ^{\frac{1}{2}} \cosh(zx)\cos(2\pi n x)dx \\ & = \left[\cosh(zx)\frac{1}{2\pi n}\sin(2\pi nx) \right]_{-\frac{1}{2}}^{\frac{1}{2}} - \int\limits_{-\frac{1}{2}} ^{\frac{1}{2}} z \sinh (zx) \frac{1}{2\pi n} \sin (2\pi nx) dx \\ & = -z \int\limits_{-\frac{1}{2}} ^{\frac{1}{2}}\sinh (zx) \frac{1}{2\pi n} \sin (2\pi nx) dx \\ & = -z \left[ \left[-\sinh (zx) \left(\frac{1}{2\pi n}\right)^2 \cos (2\pi nx) \right]_{-\frac{1}{2}}^{\frac{1}{2}} + z \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \cosh (zx)\left(\frac{1}{2\pi n}\right)^2 \cos (2\pi nx) dx \right] \end{align*}

$$ \Rightarrow \left(1 + \left(\frac{z}{2\pi n}\right)^2\right)\int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \cosh (zx) \cos (2\pi nx) dx = 2z\left(\frac{z}{2\pi n}\right)^2 (-1)^n \sinh (\frac{z}{2}) $$

$$ \rightarrow \alpha_n = \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \cosh (zx) \cos (2\pi nx) dx = \frac{2z}{(2\pi n)^2 + z^2} (-1)^n \sinh (\frac{z}{2}) $$ Where we use $$\cos(2\pi n x)$$ because the integrand is even. Therefore we get for the Fourieseries of g:

$$ \cosh (zx) = \frac{2}{z} \sinh (\frac{z}{2}) \left[ 1 + \sum\limits_{n \in \mathbb{Z} \\ n \neq 0} (-1)^n \frac{2z}{4{\pi}^2 n^2 + z^2} \cos (2\pi n x) \right] $$ $$ = \frac{2}{z} \sinh(\frac{z}{2}) \left[ 1 + 2\sum\limits_{n=1}^\infty (-1)^n \frac{2z}{4{\pi}^2 n^2 + z^2} \cos (2\pi n x) \right]$$

Now we chose $$x = \frac{1}{2}$$:

$$ \cosh (\frac{z}{2}) = \frac{2}{z} \sinh(\frac{z}{2}) \left[ 1 + 2\sum\limits_{n=1}^\infty (-1)^n \frac{z^2}{4\pi^2 n^2} \frac{1}{1+\frac{z^2}{4\pi^2 n^2}} \right] $$

$$ \underbrace{\frac{z}{2} \frac{\cosh (\frac{z}{2})}{\sinh(\frac{z}{2})}}_{f(z)} = 1 + 2\sum\limits_{n=1}^\infty (-1)^n \frac{z^2}{4{\pi}^2 n^2} \frac{1}{1+\frac{z^2}{4{\pi}^2 n^2}} $$ $$ = 1 + \sum\limits_{n=1}^\infty \sum\limits_{k=0}^\infty \frac{2z^2}{4{\pi}^2 n^2} \left(-\frac{z^2}{4\pi^2 n^2}\right)^k$$

Because both series are absolute convergent and uniform convergent we get after some calculation: $$ f(z) = 1 + \sum\limits_{k = 1}^\infty 2 (-1)^{k-1} \zeta(2k) \frac{z^{2k}}{(2\pi)^{2k}} $$ but above we also calculated

$$ f(z) =1 + \sum\limits_{n=1}^\infty B_{2n} \frac{z^{2n}}{(2n)!} $$ With a coefficient comparison we get:

$$ \zeta(2n) = - \frac{B_{2n}(-1)^{2n}(2\pi)^{2n}}{2(2n)!} = -\frac{(2\pi i)^{2n}}{2(2n)!}B_{2n} \quad\quad\quad \forall n \in \mathbb{N} $$ which is what we had to show.