$\sum \limits_{n=1}^{\infty} \left(\zeta(2n) -\beta(2n) \right)$

Prove that:

$$\sum_{n=1}^{\infty}\left( \zeta(2n)-\beta(2n) \right)=\frac{1}{2}+\frac{\ln 2}{2}$$

Solution

Using absolute convergence of the given series we have successively:

\begin{align*}
\sum_{n=1}^{\infty} \left ( \zeta(2n)-\beta(2n) \right ) &=\sum_{n=1}^{\infty} \left [ \sum_{m=1}^{\infty} \frac{1}{m^{2n}} - \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{\left ( 2m-1 \right )^{2n}} \right ] \\
 &=\sum_{n=1}^{\infty} \left [ \sum_{m=2}^{\infty} \frac{1}{m^{2n}} - \sum_{m=2}^{\infty} \frac{(-1)^{m-1}}{\left ( 2m-1 \right )^{2n}}  \right ] \\
 &= \sum_{m=2}^{\infty} \left [ \sum_{n=1}^{\infty} \frac{1}{m^{2n}} - \sum_{n=1}^{\infty} \frac{(-1)^{m-1}}{\left ( 2m-1 \right )^{2n}}  \right ]\\
 &= \sum_{m=2}^{\infty} \left [ \frac{1}{(m+1)(m-1)} + \frac{(-1)^m}{(2m-1)^2 } \cdot \frac{1}{1-\frac{1}{(2m-1)^2}} \right ]\\
 &= \sum_{m=2}^{\infty}\left [ \frac{1}{(m+1)(m-1)} + \frac{(-1)^m}{\left ( 2m-1 \right )^2-1} \right ] \\
 &=\sum_{m=2}^{\infty} \left [ \frac{1}{(m+1)(m-1)} + \frac{(-1)^m}{4(m-1)m} \right ] \\
 &=\frac{1}{2}\sum_{m=2}^{\infty}\left [ \frac{1}{m-1} -\frac{1}{m+1}  \right ] +\frac{1}{4}\sum_{m=2}^{\infty} \left [ \frac{(-1)^m}{m-1}  - \frac{(-1)^m}{m} \right ] \\
 &=\frac{3}{4} +\frac{2\ln 2 -1}{4} \\
 &=\frac{1}{2} + \frac{\ln 2}{2}
\end{align*}