A closed form of $\sum_{n=1}^\infty\left[ H_n^2-\left(\ln n+\gamma+\frac1{2n} \right)^2\right]$ The series of squares of harmonic numbers
$$
\sum_{n=1}^\infty H_n^2 \tag1
$$ is divergent since $\displaystyle \lim_{n \to \infty} H_n^2 \ne 0$, actually from the classic result (6.3.18),
$$
H_n=\ln n+\gamma+\frac1{2n}+O\left(\frac1{n^2}\right),\qquad  \, n \to \infty,
$$ where $\gamma$ is the Euler-Mascheroni constant, one gets as $n \to \infty$, 
$$
H_n^2=\left(\ln n+\gamma+\frac1{2n} \right)^2+O\left(\frac{\ln n}{n^2}\right)\tag2
$$which tends to infinity. 
Then the following new series

$$
\sum_{n=1}^\infty \color{grey}{\left[\color{#151515}{\: H_n^2-\left(\ln n+\gamma+\frac1{2n} \right)^2}\: \right]} \tag3
$$

may be seen as a sort of regularization of $(1)$. The series $(3)$ is absolutely convergent as one may directly deduce from the comparison test with a Bertrand series, using $(2)$.
Question. What is a closed form of $(3)$? 
My intuition is that $(3)$ admits a closed form in terms of known constants (or here).  I've used the advanced Inverse Symbolic Calculator ISC $2.0$ but it found nothing. My recent attempt, not yet fruitful, has been to convert $(3)$ into an integral representation, starting  with
$$
\begin{align}
-\int_0^1\!\left(\!\frac1{\ln x}+\frac1{1-x}-\frac12\!\right)\!x^{n-1}\:dx&=H_n-\ln n-\gamma-\frac1{2n},\quad n\ge1,
\end{align}
$$ and trying to employ similar techniques used here. 
Analogous considerations are here or here, one may also explore some variations of $(3)$, like
$$
\sum_{n=1}^\infty \color{grey}{\left[\color{#151515}{\: H_n^q-\left(\ln n+\gamma+\frac1{2n} \right)^q}\: \right]}, \,\sum_{n=1}^\infty (-1)^n \!\color{grey}{\left[\color{#151515}{\: H_n^q-\left(\ln n+\gamma+\frac1{2n} \right)^q}\: \right]}.
$$
 A: The given series admits a closed form.

Proposition.$$
\sum_{n=1}^\infty \color{grey}{\left[\color{#151515}{\: H_n^2-\left(\ln n+\gamma+\frac1{2n} \right)^2}\: \right]}=\frac12\ln^2(2\pi)-\gamma\ln (2\pi)-\frac12\gamma^2-2\gamma_1-1. \qquad (\star)
$$

where $\gamma_1$ is a Stieltjes constant.
(Sketch of a proof).
Let us consider, for $a\ge 0$,
$$
S(a):=\sum_{n=1}^\infty \color{grey}{\left[\color{#151515}{\: \left(\psi(n+a+1)+\gamma\right)^2-\left(\ln (n+a)+\gamma+\frac1{2(n+a)} \right)^2}\: \right]}, \tag1
$$
where throughout  $\displaystyle \psi :=\left(\text{Log}\: \Gamma  \right)'$ is the digamma function. From the standard identity
$$
\psi(n+1)+\gamma=H_n\qquad n=1,2,\cdots,\tag2
$$
we have
$$
S(0)=\sum_{n=1}^\infty \color{grey}{\left[\color{#151515}{\: H_n^2-\left(\ln n+\gamma+\frac1{2n} \right)^2}\: \right]}. \tag3
$$
One is allowed to differentiate $S(a)$ termwise obtaining
$$
\begin{align}
S'(a)=\sum_{n=1}^\infty &\left[2\:\psi'(n+a+1)\left(\psi(n+a+1)+\gamma\right)\color{#FFFFFF}{\frac2{(n+a)}}\right.
\\&-\left.\left(\frac2{(n+a)}-\frac1{(n+a)^2}\right)\!\left(\ln (n+a)+\gamma+\frac1{2(n+a)} \right)\right], \tag4
\end{align}
$$ then we are lead to consider the partial sum,
$$
\begin{align}
S_N'(a)=&\sum_{n=1}^N 2\:\psi'(n+a+1)\left(\psi(n+a+1)+\gamma\right)
\\-&\sum_{n=1}^N\left(\frac2{(n+a)}-\frac1{(n+a)^2}\right)\left(\ln (n+a)+\gamma+\frac1{2(n+a)} \right).\tag5
\end{align}
$$ We have proved here that $\displaystyle \sum_{n=1}^N \psi'(n)\psi(n)$ has a closed form, this result can be extended as follows.

Theorem. Let $a$ be any non-negative real number. We have
  $$
\begin{align}
&\sum_{n=1}^N \psi'(n+a+1)\psi(n+a+1)
\\&=\left((N+a+1) \psi(N+a+2)-(a+1)\psi(a+2)\right)' \psi(N+a+1)
\\\\&-\left((N+a+1) \psi(N+a+2)-(a+2)\psi(a+3)\right)' 
\\\\&+\left((a+1)\psi(a+2)\right)'\left(\psi(N+a+1)-\psi(a+2)\right)
 \tag6
\\\\&+\frac12 \psi'(N+a+1)-\frac12\psi'(a+2)-\frac12 \left(\psi'(N+a+1)\right)^2+\frac12 \left(\psi'(a+2)\right)^2.
\end{align}
$$ 

Proof. One uses a summation by parts with
$$
f_n(a)=\psi(n+a+1),\quad g_n(a)=\psi'(n+a+1),\quad n\ge1,
$$ taking into account that
$$
\begin{align}
&\sum_{n=1}^N \psi'(n+a+1)=\left((N+a+1) \psi(N+a+2)-(a+1)\psi(a+2)\right)'. \tag7
\end{align}
$$ 
Then the second sum on the right hand side of $(5)$ satisfies
$$
\begin{align}
&\sum_{n=1}^N\left(\frac2{(n+a)}-\frac1{(n+a)^2}\right)\left(\ln (n+a)+\gamma+\frac1{2(n+a)}\right)
\\\\&=2\gamma_1(a+1)-2\gamma_1(N+a+1)+2\gamma \psi(N+a+1)-2\gamma\psi(a+1)
 \tag8
\\\\&+\frac14 \psi''(N+a+1)-\frac14\psi''(a+2)+\gamma_1'(a+1)-\gamma_1'(N+a+1)
\\\\&+(\gamma+1)\psi'(N+a+1)-(\gamma+1)\psi'(a+1),
\end{align}
$$
where we have used the generalized Stieltjes constant,
$$
\begin{align}
\gamma_1(a+1)=\lim_{N \to \infty}\left(\sum_{n=1}^N\frac{\ln(n+a)}{n+a}-\frac12\:\ln^2 \left(N+a\right)\right).
\end{align}
$$ 
We then insert $(6)$, $(7)$ and $(8)$ into $(5)$ and let $N \to \infty$ to get
$$
\begin{align}
S'(a)=&\:\left(2a-2(a+2)\psi(a+3)\right)'+\left(2\gamma a-2\gamma(a+1)\psi(a+2)\right)'
\\\\-&\:\left(\psi(a+2)-(a+1) \left(\psi(a+2)\right)^2\right)'-2\gamma_1(a+1)+2\gamma\psi(a+1)
 \tag9
\\\\+&\frac14\psi''(a+2)-\gamma_1'(a+1)+(\gamma+1)\psi'(a+1).
\end{align}
$$
Finally, integrating $(9)$ using
$$
2\int_1^{1+a}\gamma_1(t)\:dt=\zeta''(0,a+1)-\zeta''(0),
$$ where $\zeta(\cdot,\cdot)$ denotes the Hurwitz zeta function and where $\zeta''(0,a+1)=\left.\partial_s^2\zeta(s,a+1)\right|_{s=0}$, determining the  constant of integration by letting $a \to \infty$ and using $(3)$ yields $(\star)$. 
A: Just some considerations for now.
I have proved here that
$$ \sum_{n=1}^{N}H_n^2 = (N+1) H_N^2-(2N+1)H_N+2N \tag{$\color{blue}{1}$} $$
and we have:
$$ \mathcal{L}^{-1}\left(\frac{\log x+\gamma}{x}\right)(s)=-\log(s)\tag{2} $$
$$ \mathcal{L}^{-1}\left(\frac{\left(\log x+\gamma\right)^2}{x}\right)(s)=-\zeta(2)+\log^2(s)\tag{3} $$
hence the partial sums of the given series can be rearranged as follows:
$$\begin{eqnarray*}\sum_{n=1}^{N}\left[H_n^2-\left(\log n+\gamma+\frac{1}{2n}\right)^2\right]&=&(\color{blue}{1})-\frac{H_N^{(2)}}{4}-\sum_{n=1}^{N}\frac{\log n+\gamma}{n}-\sum_{n=1}^{N}n\frac{(\log n+\gamma)^2}{n}\\&=&(\color{blue}{1})-\frac{H_N^{(2)}}{4}+\int_{0}^{+\infty}\frac{\log(s)(1-e^{-Ns})}{e^s-1}\,ds\\&-&\int_{0}^{+\infty}\frac{e^{-N s} \left(e^{(1+N) s}+N-e^s (1+N)\right)\left(\zeta(2)-\log^2 s\right)}{\left(-1+e^s\right)^2}\,ds\end{eqnarray*}$$
If we find a way to distribute $(\color{blue}{1})$ over the last two integrals, in a way ensuring they are convergent integrals as $N\to +\infty$, we are done. At first sight the closed form of the LHS appears to be related with $\zeta(2)$, $\zeta'(0)=-\frac{1}{2}\log(2\pi)$ and
$$ \zeta''(0)=\frac{\gamma^2}{2}-\frac{\pi ^2}{24}-\frac{1}{2}\log^2(2\pi)+\gamma_1$$
with $\gamma_1$ being a Stieltjes constant. An alternative, brute-force way is just to compute the asymptotic expansions of
$$ \sum_{n=1}^{N}\log^2(n),\qquad \sum_{n=1}^{N}\frac{\log(n)}{n} $$
with the sufficient degree of accuracy (I guess that to stop at the $O\left(\frac{1}{N^3}\right)$ term is enough), that is just a tedious exercise about summation by parts. 
