Nature of constant $c$ from $\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{n}(1/k)^{1/k}-n+\frac{\log^2(n)}{2}-1\right)=c$

Question

If we take $$\sum\limits_{k=1}^{n}(1/k)^{1/k}=a(n)$$ so $$\lim\limits_{n\to\infty}\left(a(n)-n+\frac{\log^2(n)}{2}-1\right)=c$$ What is the nature of constant $c$? Is it really constant (maybe it function or sum of function and constant)?

Answer 1

Computed it using PARI/GP (this is not an answer to the question on "nature"...). Write $$1+c=\sum_{n=1}^{\infty}\left((1/n)^{1/n}-1+\frac{\log^2(n+1)-\log^2n}{2}\right)=F+G$$ (PARI fails to compute it in this form), where $F=\displaystyle\sum_{n=1}^{\infty}f(n)$, $G=\displaystyle\sum_{n=1}^{\infty}g\Big(\frac{\log n}{n}\Big)$, $$f(z)=\frac{\log^2(1+z)-\log^2z}{2}-\frac{\log z}{z},\quad g(z)=e^{-z}-1+z.$$ Here, PARI's sumnum and sumnumap compute $G$ correctly, but not $F$ (which is just $-\gamma_1$ incidentally; see Stieltjes constants). It seems that in $$F=\frac{f(1)}{2}+\int_1^\infty f(x)\,dx+i\int_0^\infty\frac{f(1+it)-f(1-it)}{e^{2\pi t}-1}\,dt$$ (this is Abel-Plana formula), PARI cannot (or I couldn't make it) compute the elementary $$\int_1^\infty f(x)\,dx=-(1-\log2)^2.$$ But, armed with these handmade interventions, I get it working:

f(z)=(log(1+z)^2-log(z)^2)/2-log(z)/z;
g(t)=imag(f(1-I*t)-f(1+I*t))/(exp(2*Pi*t)-1);
a(n)={my(r=log(n)/n);return(exp(-r)-1+r)};
b=(log(2)/2)^2-(1-log(2))^2-1;
b+sumnum(n=1,a(n))+intnum(t=0,[+oo,2*Pi],g(t))

gives $0.0090071648881637765017784106780358074$ (with the default precision).