Vtyjkyuk

I did my best to get the notation correct but it is likely wrong so let me explain. The sum on the right tends to $fracpi²6$. As it is incrementally getting closer to $fracpi²6$ I want to see what the delta is to $fracpi²6$ and I would like to sum those deltas. I noticed this sum is a very close approximation to $ln(x)$ but it tends to be off by ~.0677. The 0677 is the limit of where my excel skills can get me but I want to find out if this actually tends to a specific number or if it goes up forever. I have gone through $x=1$ to $1,000,000$. Is there a way to put this into wolfram alpha?

$$ln(x)-displaystylesum_x=1^infty( fracpi²6 - displaystylesum_x=1^infty frac1x²)=?$$

Your notation does not make much sense, but trying to interpolate your actual question you seem to ask about the value of the limit
$$ lim_nto +inftyleft[log(n)-sum_k=1^nleft(zeta(2)-H_k^(2)right)right]$$
where by summation by parts
$$ sum_k=1^nH_k^(2) = n H_n^(2)-sum_k=1^n-1frack(k+1)^2=(n+1) H_n^(2)-H_n $$
such that
$$ sum_k=1^nleft(zeta(2)-H_k^(2)right) = nleft(zeta(2)-H_n^(2)right)-H_n^(2)+H_n. $$
Now $lim_nto +inftyH_n^(2)=zeta(2)$ and $lim_nto +inftyleft(H_n-log(n)right)=gamma$. By Cesàro-Stolz
$$ lim_nto +inftyfraczeta(2)-H_n^(2)frac1n=lim_nto +inftyfracfrac1(n+1)^2frac1n(n+1)=1$$
hence
$$ lim_nto +inftyleft[log(n)-sum_k=1^nleft(zeta(2)-H_k^(2)right)right]=colorredzeta(2)-gamma-1approx 0.0677184.$$

Using Jack D'Aurizio's solution, we end with
$$S_n=log(n)+H_n+1-(n+1), H_n+1^(2)+fracpi ^2 6n$$ Using the asymptotics of harmonic numbers, this leads to
$$S_n=left(fracpi ^26-gamma-1right)-frac1n+frac512
n^2+Oleft(frac1n^3right)$$ which shows the limit and how it is approached.

Moreover, this gives a quite good approximation of $S_n$; for example,
$$S_10=frac3586319254016-frac5 pi ^23+log (10)approx -0.0282788$$ while the above expansion leads to
$$-frac263240-gamma +fracpi ^26approx -0.0281149$$