バーゼル問題の解決その後ζ(4)

$$ \quad \sum_{k=1}^{n} \frac{1}{ \tan^4{ \frac{k \pi}{2n+1} } }= \lbrace \sum_{k=1}^{n} \frac{1}{ \tan^2{ \frac{k \pi}{2n+1} } } \rbrace^2-2\sum_{j< k}^{} \lbrace \frac{1}{ \tan^2{ \frac{j \pi}{2n+1} }}\frac{1}{ \tan^2{ \frac{k \pi}{2n+1} }}\rbrace $$
「バーゼル問題の解決」のpostから，
$$ \quad \sum_{j< k}^{} \lbrace \frac{1}{ \tan^2{ \frac{j \pi}{2n+1} }}\frac{1}{ \tan^2{ \frac{k \pi}{2n+1} }}\rbrace $$
は，「解と係数の関係」から，
$$ \quad \sum_{r=0}^{n} {}_{2n+1} \mathrm{ C }_{2r}{(-1)^{n-r}x^r} $$
の$x^{n-2}$の係数を考えて，
$$ \quad \sum_{j< k}^{} \lbrace \frac{1}{ \tan^2{ \frac{j \pi}{2n+1} }}\frac{1}{ \tan^2{ \frac{k \pi}{2n+1} }}\rbrace=\frac{{}_{2n+1} \mathrm{ C }_{2n-4}}{{}_{2n+1} \mathrm{ C }_{2n}}$$
$$ \quad \sum_{j< k}^{} \lbrace \frac{1}{ \tan^2{ \frac{j \pi}{2n+1} }}\frac{1}{ \tan^2{ \frac{k \pi}{2n+1} }}\rbrace= \frac{n(n-1)(2n-1)(2n-3)}{30} $$
よって，
$$ \quad \sum_{k=1}^{n} \frac{1}{ \tan^4{ \frac{k \pi}{2n+1} } }= \lbrace \frac{n(2n-1)}{3} \rbrace^2-\frac{n(n-1)(2n-1)(2n-3)}{15} $$
$$ \quad \sum_{k=1}^{n} (\frac{1}{ \tan^2{ \frac{k \pi}{2n+1} } }+1)^2= \sum_{k=1}^{n} \frac{1}{ \tan^4{ \frac{k \pi}{2n+1} } }+ \frac{2n(2n-1)}{3}+n $$
ここで，
$$ \quad \frac{1}{(2n+1)^4} \sum_{k=1}^{n} \frac{1}{ \tan^4{ \frac{k \pi}{2n+1} } } <\frac{1}{ \pi ^4} \sum_{k=1}^{n} \frac{1}{ k^4 } <\frac{1}{(2n+1)^4} \sum_{k=1}^{n} (\frac{1}{ \tan^2{ \frac{k \pi}{2n+1} } }+1)^2$$
極限は，
$$ \quad \frac{1}{(2n+1)^4} \sum_{k=1}^{n} \frac{1}{ \tan^4{ \frac{k \pi}{2n+1} } } \rightarrow \frac{1}{36}- \frac{1}{60}= \frac{1}{90} $$

$$ \quad \frac{1}{(2n+1)^4} \sum_{k=1}^{n} (\frac{1}{ \tan^2{ \frac{k \pi}{2n+1} } }+1)^2\rightarrow \frac{1}{90}+0+0=\frac{1}{90} $$
よって，「ハサミウチの原理」から，
$$ \quad \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{ k^4 } = \frac{ \pi^4}{90} ． $$