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The Dirichlet beta function is defined as $$ \beta(2m+1)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}\tag{1} $$ Equations $(8)$ and $(9)$ in this answer say that $\beta(1)=\frac\pi4$ and $$ \beta(2m+1) = -\sum_{k=1}^m \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2m-2k+1)\tag{2} $$ If we reindex recursion $(7)$ derived below, we get that the even Euler numbers are defined by $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2m}=-\sum_{k=1}^m\binom{2m}{2k}\mathrm{E}_{2m-2k}\tag{3} $$ then notice that $(2)$ is the same as $(3)$ if we set $$ \beta(2m+1)=\frac{(-1)^m\mathrm{E}_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}\tag{4} $$ QED

Recursion for the even Euler numbers

The Exponential Generating Function for the Euler numbers is $\mathrm{sech}(x)$. This means that the odd Euler numbers are $0$ and $$ \mathrm{sech}(x)=\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\tag{5} $$ Therefore, $$ \begin{align} 1 &=\cosh(x)\,\mathrm{sech}(x)\\[9pt] &=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{6} \end{align} $$ Equation $(6)$ says that $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}\tag{7} $$

The Dirichlet beta function is defined as $$ \beta(2m+1)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}\tag{1} $$ Equations $(8)$ and $(9)$ in this answer say that $\beta(1)=\frac\pi4$ and $$ \beta(2m+1) = -\sum_{k=1}^m \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2m-2k+1)\tag{2} $$ If we reindex recursion $(7)$ derived below, we get that the even Euler numbers are defined by $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2m}=-\sum_{k=1}^m\binom{2m}{2k}\mathrm{E}_{2m-2k}\tag{3} $$ then notice that $(2)$ is the same as $(3)$ if we set $$ \beta(2m+1)=\frac{(-1)^m\mathrm{E}_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}\tag{4} $$ QED

Recursion for the even Euler numbers

The Exponential Generating Function for the Euler numbers is $\mathrm{sech}(x)$. This means that the odd Euler numbers are $0$ and $$ \mathrm{sech}(x)=\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\tag{5} $$ Therefore, $$ \begin{align} 1 &=\cosh(x)\,\mathrm{sech}(x)\\[9pt] &=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{6} \end{align} $$ Equation $(6)$ says that $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}\tag{7} $$

The Dirichlet beta function is defined as $$ \beta(2m+1)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}\tag{1} $$ Equations $(8)$ and $(9)$ in this answer say that $\beta(1)=\frac\pi4$ and $$ \beta(2m+1) = -\sum_{k=1}^m \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2m-2k+1)\tag{2} $$ If we reindex recursion $(7)$ derived below, we get that the even Euler numbers are defined by $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2m}=-\sum_{k=1}^m\binom{2m}{2k}\mathrm{E}_{2m-2k}\tag{3} $$ then notice that $(2)$ is the same as $(3)$ if we set $$ \beta(2m+1)=\frac{(-1)^m\mathrm{E}_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}\tag{4} $$ QED

Recursion for the even Euler numbers

The Exponential Generating Function for the Euler numbers is $\mathrm{sech}(x)$. This means that the odd Euler numbers are $0$ and $$ \mathrm{sech}(x)=\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\tag{5} $$ Therefore, $$ \begin{align} 1 &=\cosh(x)\,\mathrm{sech}(x)\\[9pt] &=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{6} \end{align} $$ Equation $(7)$ says that $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}\tag{7} $$

The Dirichlet beta function is defined as $$ \beta(2m+1)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}\tag{1} $$ Equations $(8)$ and $(9)$ in this answer say that $\beta(1)=\frac\pi4$ and $$ \beta(2m+1) = -\sum_{k=1}^m \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2m-2k+1)\tag{2} $$ If we reindex recursion $(7)$ derived below, we get that the even Euler numbers are defined by $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2m}=-\sum_{k=1}^m\binom{2m}{2k}\mathrm{E}_{2m-2k}\tag{3} $$ then notice that $(2)$ is the same as $(3)$ if we set $$ \beta(2m+1)=\frac{(-1)^m\mathrm{E}_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}\tag{4} $$ QED

Recursion for the even Euler numbers

The Exponential Generating Function for the Euler numbers is $\mathrm{sech}(x)$. This means that the odd Euler numbers are $0$ and $$ \mathrm{sech}(x)=\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\tag{5} $$ Therefore, $$ \begin{align} 1 &=\cosh(x)\,\mathrm{sech}(x)\\[9pt] &=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{6} \end{align} $$ Equation $(6)$ says that $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}\tag{7} $$

The Dirichlet beta function is defined as $$ \beta(2m+1)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}\tag{1} $$ Equation $(9)$ in this answer says that $\beta(1)=\frac\pi4$ and $$ \beta(2m+1) = -\sum_{k=1}^m \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2m-2k+1)\tag{2} $$ So what we need to show is that $$ \frac{(-1)^m\mathrm{E}_{2m}\pi^{2m+1}}{4^{m+1}(2m)!} =-\sum_{k=1}^m\frac{(-\pi^2/4)^k}{(2k)!}\;\frac{(-1)^{m-k}\mathrm{E}_{2m-2k}\pi^{2m-2k+1}}{4^{m-k+1}(2m-2k)!}\tag{3} $$ which is equivalent to $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2m}=-\sum_{k=0}^{m-1}\binom{2m}{2k}\mathrm{E}_{2k}\tag{4} $$ which is the recursion for the even Euler numbers.

Recursion for the even Euler numbers

The Exponential Generating Function for the Euler numbers is $\mathrm{sech}(x)$. This means that the odd Euler numbers are $0$ and $$ \mathrm{sech}(x)=\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\tag{5} $$ Therefore, $$ \begin{align} 1 &=\cosh(x)\,\mathrm{sech}(x)\\[9pt] &=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{6} \end{align} $$ Equation $(6)$ says that $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}\tag{7} $$

The Dirichlet beta function is defined as $$ \beta(2m+1)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}\tag{1} $$ Equations $(8)$ and $(9)$ in this answer say that $\beta(1)=\frac\pi4$ and $$ \beta(2m+1) = -\sum_{k=1}^m \frac{(-\pi^2/4)^k}{(2k)!}\;\beta(2m-2k+1)\tag{2} $$ If we reindex recursion $(7)$ derived below, we get that the even Euler numbers are defined by $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2m}=-\sum_{k=1}^m\binom{2m}{2k}\mathrm{E}_{2m-2k}\tag{3} $$ then notice that $(2)$ is the same as $(3)$ if we set $$ \beta(2m+1)=\frac{(-1)^m\mathrm{E}_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}\tag{4} $$ QED

Recursion for the even Euler numbers

The Exponential Generating Function for the Euler numbers is $\mathrm{sech}(x)$. This means that the odd Euler numbers are $0$ and $$ \mathrm{sech}(x)=\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\tag{5} $$ Therefore, $$ \begin{align} 1 &=\cosh(x)\,\mathrm{sech}(x)\\[9pt] &=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\sum_{n=0}^\infty\frac{\mathrm{E}_{2n}}{(2n)!}x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac1{(2n-2k)!}\frac{\mathrm{E}_{2k}}{(2k)!}\right)x^{2n}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{2n}{2k}\mathrm{E}_{2k}\right)\frac{x^{2n}}{(2n)!}\tag{6} \end{align} $$ Equation $(7)$ says that $\mathrm{E}_0=1$ and $$ \mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}\tag{7} $$