Taylor series expansion of the cotangent function and denominators of the Riemann zeta function at positive even integers

Question

Consider the Taylor series expansion of the cotangent function:

$$ \frac{1}{x} - \frac{x}{3} - \frac{x^3}{45} - \frac{2 x^5}{945} - \frac{x^7}{4725} - \frac{2 x^9}{93555} - \frac{1382 x^{11}}{638512875} - \frac{4 x^{13}}{18243225} + O(x^{15}). $$

Now let $\zeta(s)$ represent the Riemann zeta function defined for positive integer $s>1$ as usual by:

$$ \zeta(s) = \sum_{n=1}^{ \infty} \frac{1}{n^s}. $$

At positive even integers $2s$ we have the identity:

$$ {\displaystyle A_{n}\zeta (2n)=B_{n}\pi ^{2n}\,\!}, $$

where the sequence $A_n$ ($A002432$ in the OEIS) represents the denominators in $\zeta(2s)$. Starting from $s=3$ these begin with:

• $\zeta (6)=1+{\frac {1}{2^{6}}}+{\frac {1}{3^{6}}}+\cdots ={\frac {\pi ^{6}}{945}}$
• $\zeta (8)=1+{\frac {1}{2^{8}}}+{\frac {1}{3^{8}}}+\cdots ={\frac {\pi ^{8}}{9450}}$
• $\zeta (10)=1+{\frac {1}{2^{{10}}}}+{\frac {1}{3^{{10}}}}+\cdots ={\frac {\pi ^{{10}}}{93555}}$
• $\zeta (12)=1+{\frac {1}{2^{{12}}}}+{\frac {1}{3^{{12}}}}+\cdots ={\frac {691\pi ^{{12}}}{638512875}}$
• $\zeta (14)=1+{\frac {1}{2^{{14}}}}+{\frac {1}{3^{{14}}}}+\cdots ={\frac {2\pi ^{{14}}}{18243225}}$

Note that the denominators of the Taylor series expansion above almost coincide (except for the first several terms) with the values of $A_n$. Below is the sequence of $A_n$, in which the numbers that appear as denominators in the Taylor expansion of $\cot x$ are underlined:

$$ 6, 90, \underline{945}, 9450, \underline{93555}, \underline{638512875}, \underline{18243225}, 325641566250, \underline{38979295480125}, \underline{1531329465290625}, \underline{13447856940643125}, \underline{201919571963756521875}, \underline{11094481976030578125}, \underline{564653660170076273671875}, \underline{5660878804669082674070015625}, \ldots $$

What explains this? Is this just a conincidence?

Answer 1

The Taylor series for the cotangent function is

$$\cot(x)=\sum_{n=0}^\infty \frac{(-1)^n 2^{2n}B_{2n}}{(2n)!}\,x^{2n-1}$$

whereas we have for $n\ge 1$

$$\zeta(2n)=\frac{(-1)^{n+1}2^{2n}B_{2n}}{2(2n)!}\,\pi^{2n}$$

So, we see that the coefficients $a_n$ of the Taylor series for the cotangent function can be written as

$$a_n=-2\frac{\zeta(2n)}{\pi^{2n}}\tag1$$

for $n\ge1$.

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Note that we can recover the partial fraction expansion representation of the cotangent function from the Taylor series with coefficients $a_n$ as given in $(1)$. Proceeding, we have

$$\begin{align} \cot(x)&=\frac1x+\sum_{n=1}^\infty a_nx^{2n-1}\\\\ &=\frac1x-2\sum_{n=1}^\infty \frac{\zeta(2n)}{\pi^{2n}}\,x^{2n-1}\\\\ &=\frac1x-\frac2x\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{k^{2n}}\left(\frac x\pi\right)^{2n}\\\\ &=\frac1x-\frac2x \sum_{k=1}^\infty \sum_{n=1}^\infty \left(\frac {x}{k\pi}\right)^{2n}\\\\ &=\frac1x-\frac2x \sum_{k=1}^\infty \frac{(x/k\pi)^2}{1-(x/k\pi)^2}\\\\ &=\frac1x+2x\sum_{k=1}^\infty \frac{1}{x^2-(k\pi)^2} \end{align}$$

as was to be shown!

Answer 2

Another way to see this: let $H(x) = \sin(x)/x$ and write $H(x)$ as an infinite product:

$$ H(x) = \prod_{n=1}^\infty \left(1-\frac{x^2}{\pi^2n^2}\right). $$

Take the negative logarithmic derivative both sides: the LHS is $$ \frac 1x - \cot(x) = \sum_{j=1}(-1)^{j-1}2^{2j}B_{2j}\frac{x^{2j-1}}{(2j)!} $$ and the RHS is $$ \sum_{n=1}^\infty\frac{2x}{n^2\pi^2}\frac 1{1-(x/n\pi)^2}. $$ Expanding $1/(1-x^2/n^2\pi^2)$ into a geometric series and rearranging gives: $$ 2\sum_{j=1}^\infty (-1)^{j-1}\zeta(2j) \frac{x^{2j-1}}{\pi^{2j}} $$