(1)

\begin{align*} \tanh^{-1}x & =\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\\ & =\frac{e^{2x}+1}{e^{2x}-1}\\ & =1+\frac{2}{e^{2x}-1}\\ & =1+\frac{1}{x}\cdot\frac{2x}{e^{2x}-1}\\ & =1+\frac{1}{x}\sum_{k=0}^{\infty}\frac{B_{k}}{k!}\left(2x\right)^{k}\\ & =1+\sum_{k=0}^{\infty}\frac{2^{k}B_{k}}{k!}x^{k-1}\\ & =1+\sum_{k=0}^{\infty}\frac{2^{2k+1}B_{2k+1}}{\left(2k+1\right)!}x^{2k}+\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{\left(2k\right)!}x^{2k-1}\\ & =1+2B_{1}+\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{\left(2k\right)!}x^{2k-1}\\ & =\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{\left(2k\right)!}x^{2k-1} \end{align*} なので、
\begin{align*} \sum_{k=0}^{\infty}\frac{T_{k}}{k!}x^{k} & =\tan x\\ & =\frac{\sin x}{\cos x}\\ & =\frac{\sin^{2}x-\cos^{2}x+\cos^{2}x}{\sin x\cos x}\\ & =\frac{\cos x}{\sin x}-\frac{\cos^{2}x-\sin^{2}x}{\sin x\cos x}\\ & =\frac{\cos x}{\sin x}-2\frac{\cos\left(2x\right)}{\sin\left(2x\right)}\\ & =\tan^{-1}x-2\tan^{-1}\left(2x\right)\\ & =i\tanh^{-1}\left(ix\right)-2i\tanh^{-1}\left(2ix\right)\\ & =i\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{\left(2k\right)!}\left(ix\right)^{2k-1}-2i\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{\left(2k\right)!}\left(2ix\right)^{2k-1}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{2^{2k}B_{2k}}{\left(2k\right)!}x^{2k-1}-2\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{2^{2k-1}2^{2k}B_{2k}}{\left(2k\right)!}x^{2k-1}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{2^{2k}\left(1-2^{2k}\right)B_{2k}}{\left(2k\right)!}x^{2k-1}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{2^{2k}\left(1-2^{2k}\right)B_{2k}}{2k}\frac{x^{2k-1}}{\left(2k-1\right)!}\\ & =\sum_{k=1}^{\infty}\left(-1\right)^{k}\frac{2^{2k}\left(1-2^{2k}\right)B_{2k}}{2k}\frac{x^{2k-1}}{\left(2k-1\right)!} \end{align*} となり、係数を比較して、
\[ \begin{cases} T_{2k-1}=\left(-1\right)^{k}\frac{2^{2k}\left(1-2^{2k}\right)B_{2k}}{2k}\\ T_{2k}=0 \end{cases} \] となる。
従って与式は成り立つ。

(2)

\begin{align*} \sum_{k=0}^{\infty}\frac{T_{k}}{k!}x^{k} & =\tan x\\ & =-i\tanh ix\\ & =-i\cosh^{-1}ix\sinh ix\\ & =-i\sum_{k=0}^{\infty}\frac{E_{2k}}{\left(2k\right)!}\left(ix\right)^{2k}\sum_{j=0}^{\infty}\frac{\left(ix\right)^{2j+1}}{\left(2j+1\right)!}\\ & =\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\left(-1\right)^{k+j}\frac{E_{2k}x^{2\left(j+k\right)+1}}{\left(2k\right)!\left(2j+1\right)!}\\ & =\sum_{j=0}^{\infty}\sum_{k=0}^{j}\left(-1\right)^{j}\frac{E_{2k}x^{2j+1}}{\left(2k\right)!\left(2j-2k+1\right)!}\\ & =\sum_{j=0}^{\infty}\left(-1\right)^{j}\sum_{k=0}^{j}C\left(2j+1,2k\right)E_{2k}\frac{x^{2j+1}}{\left(2j+1\right)!}\\ & =\sum_{j=1}^{\infty}\left(-1\right)^{j-1}\sum_{k=0}^{j-1}C\left(2j-1,2k\right)E_{2k}\frac{x^{2j-1}}{\left(2j-1\right)!} \end{align*} 係数を比較して、
\[ \begin{cases} T_{2k-1}=\left(-1\right)^{k-1}\sum_{j=0}^{k-1}C\left(2k-1,2j\right)E_{2j}\\ T_{2k}=0 \end{cases} \] となる。
従って与式は成り立つ。

(3)

\begin{align*} \left(-1\right)^{k-1}\sum_{j=0}^{k-1}C\left(2k-1,2j\right)E_{2j} & =T_{2k-1}\\ & =\left(-1\right)^{k}\frac{2^{2k}\left(1-2^{2k}\right)B_{2k}}{2k} \end{align*} なので、
\begin{align*} B_{2k} & =\left(-1\right)^{k}\frac{2k}{2^{2k}\left(1-2^{2k}\right)}\left(-1\right)^{k-1}\sum_{j=0}^{k-1}C\left(2k-1,2j\right)E_{2j}\\ & =\frac{2k}{2^{2k}\left(2^{2k}-1\right)}\sum_{j=0}^{k-1}C\left(2k-1,2j\right)E_{2j}\\ & =\frac{2k}{2^{2k}\left(2^{2k}-1\right)}\sum_{j=0}^{k-1}C\left(2k-1,2j\right)E_{2j} \end{align*} となる。
従って与式は成り立つ。

(4)

(3)より、
\begin{align*} E_{2n} & =\sum_{k=1}^{n}\frac{2^{2k}\left(1-2^{2k}\right)}{2k}C\left(2n,2k-1\right)B_{2k}+1\\ & =\sum_{k=1}^{n}\frac{2^{2k}}{2k}C\left(2n,2k-1\right)B_{2k}-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+1\\ & =\sum_{k=1}^{n}\frac{2^{2k}\left(2n\right)!B_{2k}}{2k\left(2k-1\right)!\left(2n-2k+1\right)!}-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+1\\ & =\left(2n\right)!\sum_{k=1}^{n}\frac{2^{2k}B_{2k}}{\left(2k\right)!\left(2n-2k+1\right)!}-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+1\\ & =\left(2n\right)!\left(\sum_{k=0}^{n}\frac{2^{2k}B_{2k}}{\left(2k\right)!\left(2n-2k+1\right)!}-\frac{B_{0}}{\left(2n+1\right)!}\right)-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+1\\ & =\left(2n\right)!\left(\frac{1}{\left(2n\right)!}-\frac{1}{\left(2n+1\right)!}\right)-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+1\cmt{\because\sum_{k=0}^{n}\frac{2^{2k}B_{2k}}{\left(2k\right)!\left(2n-2k+1\right)!}=\frac{1}{\left(2n\right)!}}\\ & =1-\frac{1}{2n+1}-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+1\\ & =-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+2-\frac{1}{2n+1} \end{align*} となるので与式は成り立つ。

(5)

(4)より、
\begin{align*} E_{2n} & =-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+2-\frac{1}{2n+1}\\ & =2\left(1-\frac{1}{2n+1}\right)-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+\frac{1}{2n+1}\\ & =2\left(2n\right)!\left(\frac{1}{\left(2n\right)!}-\frac{1}{\left(2n+1\right)!}\right)-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+\frac{1}{2n+1}\\ & =2\left(2n\right)!\left(\sum_{k=0}^{n}\frac{2^{2k}B_{2k}}{\left(2k\right)!\left(2n-2k+1\right)!}-\frac{B_{0}}{\left(2n+1\right)!}\right)-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+\frac{1}{2n+1}\cmt{\sum_{k=0}^{n}\frac{2^{2k}B_{2k}}{\left(2k\right)!\left(2n-2k+1\right)!}=\frac{1}{\left(2n\right)!}}\\ & =2\left(2n\right)!\sum_{k=1}^{n}\frac{2^{2k}B_{2k}}{\left(2k\right)!\left(2n-2k+1\right)!}-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+\frac{1}{2n+1}\\ & =2\left(2n\right)!\sum_{k=1}^{n}\frac{2^{2k}B_{2k}}{2k\left(2k-1\right)!\left(2n-2k+1\right)!}-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+\frac{1}{2n+1}\\ & =\sum_{k=1}^{n}\frac{2^{2k}C\left(2n,2k-1\right)B_{2k}}{k}-\sum_{k=1}^{n}\frac{2^{2k}2^{2k-1}}{k}C\left(2n,2k-1\right)B_{2k}+\frac{1}{2n+1}\\ & =\sum_{k=1}^{n}\frac{2^{2k}\left(1-2^{2k-1}\right)C\left(2n,2k-1\right)B_{2k}}{k}+\frac{1}{2n+1} \end{align*}

(6)

(4)より、
\begin{align*} E_{2n} & =-\sum_{k=1}^{n}\frac{\left(2^{2k}\right)^{2}}{2k}C\left(2n,2k-1\right)B_{2k}+2-\frac{1}{2n+1}\\ & =-\frac{1}{2n+1}\sum_{k=1}^{n}\left(2^{2k}\right)^{2}C\left(2n+1,2k\right)B_{2k}+2-\frac{1}{2n+1}\\ & =-\frac{1}{2n+1}\sum_{k=0}^{n}\left(2^{2k}\right)^{2}C\left(2n+1,2k\right)B_{2k}+2\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n}\left(2^{k}\right)^{2}C\left(2n+1,k\right)B_{k}+\frac{1}{2n+1}\sum_{k=0}^{n-1}\left(2^{2k+1}\right)^{2}C\left(2n+1,2k+1\right)B_{2k+1}+2\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n}\left(2^{k}\right)^{2}C\left(2n+1,k\right)B_{k}+\frac{1}{2n+1}\sum_{k=0}^{n-1}\left(2^{2k+1}\right)^{2}C\left(2n+1,2k+1\right)\left(-\frac{1}{2}\delta_{0,k}\right)+2\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n}\left(2^{k}\right)^{2}C\left(2n+1,k\right)B_{k}-\frac{1}{2n+1}\sum_{k=0}^{n-1}2^{2}C\left(2n+1,1\right)\frac{1}{2}\delta_{0,k}+2\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n}\left(2^{k}\right)^{2}C\left(2n+1,k\right)B_{k}-2H_{1}\left(n-1\right)+2\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n}\left(2^{k}\right)^{2}C\left(2n+1,k\right)B_{k}+2\left(1-H_{1}\left(n-1\right)\right)\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n}2^{2k}C\left(2n+1,k\right)B_{k}+2\delta_{0,n} \end{align*} 更に計算を進めると、
\begin{align*} E_{2n} & =-\frac{1}{2n+1}\sum_{k=0}^{2n}2^{2k}C\left(2n+1,k\right)B_{k}+2\delta_{0,n}\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n+1}2^{2k}C\left(2n+1,k\right)B_{k}+2^{2\left(2n+1\right)}C\left(2n+1,2n+1\right)B_{2n+1}+2\delta_{0,n}\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n+1}2^{2k}C\left(2n+1,k\right)B_{k}+2^{4n+2}B_{2n+1}+2\delta_{0,n}\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n+1}2^{2k}C\left(2n+1,k\right)B_{k}+2^{4n+2}\left(-\frac{1}{2}\delta_{0,n}\right)+2\delta_{0,n}\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n+1}2^{2k}C\left(2n+1,k\right)B_{k}-2\delta_{0,n}+2\delta_{0,n}\\ & =-\frac{1}{2n+1}\sum_{k=0}^{2n+1}2^{2k}C\left(2n+1,k\right)B_{k} \end{align*} となるので与式は成り立つ。

(7)

(1),(2)より、
\(k\in\mathbb{N}\)のとき、
\begin{align*} \left(-1\right)^{k}\frac{2^{2k}\left(1-2^{2k}\right)B_{2k}}{2k} & =T_{2k-1}\\ & =\left(-1\right)^{k-1}\sum_{j=0}^{k-1}C\left(2k-1,2j\right)E_{2j} \end{align*} なので、
\[ B_{2k}=\frac{2k}{2^{2k}\left(2^{2k}-1\right)}\sum_{j=0}^{k-1}C\left(2k-1,2j\right)E_{2j} \] となり、与式が成り立つ。