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 $\frac{\pi²}{6}$. As it is incrementally getting closer to $\frac{\pi²}{6}$ I want to see what the delta is to $\frac{\pi²}{6}$ 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?
$$\ln(x)-\displaystyle\sum_{x=1}^{\infty}( \frac{\pi²}{6} - \displaystyle\sum_{x=1}^{\infty} \frac{1}{x²})=?$$
Your notation does not make much sense, but trying to interpolate your actual question you seem to ask about the value of the limit $$ \lim_{n\to +\infty}\left[\log(n)-\sum_{k=1}^{n}\left(\zeta(2)-H_k^{(2)}\right)\right]$$ where by summation by parts $$ \sum_{k=1}^{n}H_k^{(2)} = n H_n^{(2)}-\sum_{k=1}^{n-1}\frac{k}{(k+1)^2}=(n+1) H_n^{(2)}-H_n $$ such that $$ \sum_{k=1}^{n}\left(\zeta(2)-H_k^{(2)}\right) = n\left(\zeta(2)-H_n^{(2)}\right)-H_n^{(2)}+H_n. $$ Now $\lim_{n\to +\infty}H_n^{(2)}=\zeta(2)$ and $\lim_{n\to +\infty}\left(H_n-\log(n)\right)=\gamma$. By Cesàro-Stolz $$ \lim_{n\to +\infty}\frac{\zeta(2)-H_n^{(2)}}{\frac{1}{n}}=\lim_{n\to +\infty}\frac{\frac{1}{(n+1)^2}}{\frac{1}{n(n+1)}}=1$$ hence $$ \lim_{n\to +\infty}\left[\log(n)-\sum_{k=1}^{n}\left(\zeta(2)-H_k^{(2)}\right)\right]=\color{red}{\zeta(2)-\gamma-1}\approx 0.0677184.$$