Euler-Mascheroni Mashup

Let us first observe the partial sums of the series, in individual parts (where we can, for now): $\displaystyle \sum_{n=1}^N H_n = (N+1)(H_{N+1} - 1)$ ; and $\displaystyle \sum_{n=1}^N \ln n = \ln N!$ . (Try proving the first as an exercise.)

As $N \to \infty$ , we find that

$\begin{aligned} \sum_{n=1}^N \left[ H_n - \gamma - \ln n \right] &= (N+1)(H_{N+1} - 1) - \gamma N - \ln N! \\ &\to (N+1)(\gamma + \ln N + \frac{1}{2N}) - \gamma N - (N\ln N - N + \dfrac{1}{2}\ln(2\pi N)) \\ &\to \frac{1}{2}(1 + \gamma - \ln(2\pi)) + \frac{1}{2}H_N \end{aligned}$

where we have used that $H_N = \gamma + \ln N + \dfrac{1}{2N} + O(N^{-2})$ , and $\ln N! = N\ln N - N + \dfrac{1}{2}\ln(2\pi N) + O(N^{-1})$ (Stirling's approximation). Therefore

$\begin{aligned} \sum_{n=1}^\infty \left[ H_n - \gamma - \ln n - \frac{1}{2n} \right] &= \lim_{N \to \infty} \sum_{n=1}^N \left[ H_n - \gamma - \ln n \right] - \frac{1}{2}H_N \\ &= \frac{1}{2}(1 + \gamma - \ln(2\pi)) \end{aligned}$

Then note that $\displaystyle \sum_{n=1}^\infty \frac{\zeta(2n)-1}{n} = \ln 2$ (see proof in comments). Finally,

$\begin{aligned} \sum_{n=1}^\infty \left[ H_n - \gamma - \ln n - \frac{\zeta(2n)}{2n} \right] &= \sum_{n=1}^\infty \left[ H_n - \gamma - \ln n - \frac{1}{2n} \right] - \sum_{n=1}^\infty \left[ \frac{\zeta(2n)-1}{2n} \right] \\ &= \frac{1}{2}(1 + \gamma - \ln(2\pi)) - \frac{1}{2}\ln 2 \\ &= \frac{1}{2}(1 + \gamma - \ln(4\pi)) \end{aligned}$