Motivation
In a recent post (Asymptotic behaviour of sums involving $k$, $\log(k)$ and $H_{k}$) I asked for the asymptotic behaviour of the sum
$$\sigma_{c}(n)=\sum_{k=1}^n H_{k} \log(k)\tag{1}$$
and I found that the constant in the asymptotic expression contained, among other known constants, the sum
$$\kappa_{c} = \sum_{m=1}^\infty \frac{B(2m)}{2m} \zeta'(2m)\\=\frac{1}{12} \zeta '(2)-\frac{\zeta '(4)}{120}+\frac{\zeta '(6)}{252}-+…\tag{2}$$
where $B(n)$ is the Bernoulli number of order $n$ and $\zeta'()$ is the derivative of the Riemann zeta function. This is a strongly divergent series, and I had to resort to the limit
$$\lim_{n\to\infty} (\sigma_{c}(n) – (\text{leading terms}))\tag{3}$$
for calculating the complete constant, a sum of known constants and $\kappa_{c}$, which, however, I could only use numerically.
Quite recently, however, in an answer to Constant term in Stirling type formula for $\sum^N_{n=1} H_n \cdot \ln(n)$, a consistent interpretation of the sum in (2) was given in terms of a convergent integral and the numerical value calculated.
Questions
This motivated me to ask for the values of similar divergent series, and for a proof of the interpretation of the following two examples:
$$\kappa_{d} :=\sum_{m=1}^\infty B(2m)\dot{=} \frac{\pi^2}{6}-\frac{3}{2}\simeq 0.144934$$
and
$$\kappa_{e} :=\sum_{m=1}^\infty \frac{B(2m)}{2m}\dot{=}\gamma -\frac{1}{2}\simeq 0.0772157 $$
Here $\gamma$ is the Euler-Mascheroni constant and $\dot{=}$ means "is understood as" in the Ramanujan sense that $1+2+3+… \dot{=} -\frac{1}{12}$
Best Answer
In both, use the following formula, available as Gradshteyn&Rhyzik 3.411.2 $$\frac{B_{2n}}{2n} = (-1)^{n+1}\,2 \int_0^\infty \frac{x^{2n-1}}{e^{2\pi\,x}-1}\,dx $$ Then a convergent integral for the second sum is derived by $$ \sum_{n=1}^\infty \frac{B_{2n}}{2n} \,\dot{=} 2 \int_0^\infty \frac{dx/x}{e^{2\pi\,x}-1}\,\sum_{n=1}^\infty (-1)^{n+1}x^{2n} =2 \int_0^\infty \frac{dx\,x}{(x^2+1)(e^{2\pi\,x}-1)}.$$ Mathematica knows this integral, although G&R 3.415.1 can also be used: $$ (A) \quad \int_0^\infty \frac{dx\,x}{(x^2+b^2)(e^{2\pi\,x}-1)} = \frac{1}{2}\big( \log{b} - \psi(b) - \frac{1}{2b} \big) $$ For $b=1$ we therefore have $$ \sum_{n=1}^\infty \frac{B_{2n}}{2n} \,\dot{=} \,\gamma - \frac{1}{2} .$$ The other sum is similar, $$ \sum_{n=1}^\infty B_{2n} \,\dot{=} 4 \int_0^\infty \frac{dx/x}{e^{2\pi\,x}-1}\,\sum_{n=1}^\infty (-1)^{n+1}n\,x^{2n} =4 \int_0^\infty \frac{dx\,x}{(x^2+1)^2(e^{2\pi\,x}-1)}.$$ Differentiate eq. $(A)$ and set $b=1$ to find $$ \sum_{n=1}^\infty B_{2n}\,\dot{=} \,\frac{\pi^2}{6} - \frac{3}{2} .$$