Vtyjkyuk

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

$$sigma_c(n)=sum_k=1^n H_k log(k)tag1$$

and I found that the constant in the asymptotic expression contained, among other known constants, the sum

$$kappa_c = sum_m=1^infty fracB(2m)2m zeta'(2m)\=frac112 zeta '(2)-fraczeta '(4)120+fraczeta '(6)252-+...tag2$$

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

$$lim_ntoinfty (sigma_c(n) - (textleading terms))tag3$$

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:

$$kappa_d :=sum_m=1^infty B(2m)dot= fracpi^26-frac32simeq 0.144934$$

and

$$kappa_e :=sum_m=1^infty fracB(2m)2mdot=gamma -frac12simeq 0.0772157 $$

Here $gamma$ is the Euler-Mascheroni constant and $dot=$ means "is understood as" in the Ramanujan sense that $1+2+3+... dot= -frac112$

In both, use the following formula, available as Gradshteyn&Rhyzik 3.411.2
$$fracB_2n2n = (-1)^n+1,2 int_0^infty fracx^2n-1e^2pi,x-1,dx $$
Then a convergent integral for the second sum is derived by
$$ sum_n=1^infty fracB_2n2n ,dot=
2 int_0^infty fracdx/xe^2pi,x-1,sum_n=1^infty (-1)^n+1x^2n
=2 int_0^infty fracdx,x(x^2+1)(e^2pi,x-1).$$
Mathematica knows this integral, although G&R 3.415.1 can also be used:
$$ (A) quad int_0^infty fracdx,x(x^2+b^2)(e^2pi,x-1) =
frac12big( logb - psi(b) - frac12b big) $$
For $b=1$ we therefore have
$$ sum_n=1^infty fracB_2n2n ,dot= ,gamma - frac12 .$$
The other sum is similar,
$$ sum_n=1^infty B_2n ,dot=
4 int_0^infty fracdx/xe^2pi,x-1,sum_n=1^infty (-1)^n+1n,x^2n
=4 int_0^infty fracdx,x(x^2+1)^2(e^2pi,x-1).$$
Differentiate eq. $(A)$ and set $b=1$ to find
$$ sum_n=1^infty B_2n,dot= ,fracpi^26 - frac32 .$$