Conjecture $\sum_{n=0}^\infty a_n= \frac{1}{2}-\frac{7 \zeta(3)}{2 \pi^2}$ Working with some integrals I stumbled upon the following slowly converging series:
$$
S =
\sum_{n = 0}^{\infty}\left(-1\right)^{n} \left[n + \frac{3}{2} + \left(n + 1\right)\left(n + 2\right) \log\left(1 - \frac{1}{n + 2}\right)\right]
$$
I have reasons to suspect that the series has a closed form:
$$S=\frac{1}{2}-\frac{7 \zeta(3)}{2 \pi^2}=0.073721601182494209 \ldots$$
The actual proof eludes me so far.

Can you prove or disprove this conjecture?

Writing the logarithm as a series we have:
\begin{align}
&\left(n + 2\right)
\log\left(1 - \frac{1}{n+2}\right) =
-\sum_{k = 1}^{\infty}
\frac{1}{k\left(n + 2\right)^{k - 1}}
\\ = &\
-1-\sum_{k=1}^\infty \frac{1}{(k+1) (n+2)^k}
\end{align}
Which turns the series into:
$$S=\sum_{n=0}^\infty (-1)^n \left(\frac{1}{2}-(n+1) \sum_{k=1}^\infty \frac{1}{(k+1) (n+2)^k} \right)$$

I can provide the way I came to this expression, but it's very long and complicated, as usual. I'd like some clear proof, if possible.

To get the sense of how slowly the series converges, for $20000$ terms the result agrees with the stated closed form in $4$ first significant digits.

The integral from which this series was obtained is (again, conjectured):
$$\int_0^1 {_2 F_1} (1,-t;2-t;-1) dt = \frac{7 \zeta(3)}{\pi^2}+\frac{1}{2}$$
I don't think it's very useful, except for numerical confirmation.
 A: For $\Re(s)> 0$, let $F(s)$ be defined as$$
\small F(s) = \sum_{n=0}^\infty (-1)^n\left[(n+1)^{1-s} + \frac 1 2 (n+1)^{-s} + \color{red}{(n+1)(n+2)}\left[(n+1)^{-s}\ln(n+1) - \color{red}{(n+2)^{-s}\ln(n+2)}\right]\right].
$$ Then $F(s)$ is analytic and $\displaystyle S= \lim_{\substack{s\to 0\\\Re(s)>0}}F(s)$ holds. Assume $\Re(s)>3$ for a moment so that each term in the summand is absolutely summable. Then by making $n+1\mapsto n$ to the red-colored term, we get
\begin{align*}
F(s) =& \sum_{n=0}^\infty (-1)^n\left[(n+1)^{1-s} + \frac 1 2 (n+1)^{-s} + 2(n+1)^{2-s} \ln(n+1)\right]\\
=&\eta(s-1) +\frac 1 2 \eta(s) -2\eta'(s-2) ,\qquad  \Re(s)>3
\end{align*} where $\eta(s)$ is the Dirichlet's eta function. By analytic continuation, this should also hold for all $\Re(s)>0$ and it follows
$$
S =  \lim_{\substack{s\to 0\\\Re(s)>0}}F(s) = \eta(-1) + \frac 1 2 \eta(0)-2\eta'(-2)=\frac 1 2 - \frac{7\zeta(3)}{2\pi^2}
$$ where the result can be derived from the functional equations satisfied by $\eta(s)$ and $\zeta(s)$, i.e.
$$
\eta(s) = (1-2^{1-s})\zeta(s),\qquad \zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s).
$$
A: Splitting up the positive and negative terms gives
$$S=\sum_{n=1}^\infty-1+2n(2n-1)\ln\left(1-\frac1{2n}\right)-2n(2n+1)\ln\left(1-\frac1{2n+1}\right)$$
Since $\displaystyle\ln\left(1-\frac1n\right)=\ln(n-1)-\ln(n)=-\int_0^1\frac1{x+n-1}~\mathrm dx$ we provide the more suitable form
$$S=\int_0^1\sum_{n=1}^\infty-1+\frac{2n(2n+1)}{x+2n}-\frac{2n(2n-1)}{x+2n-1}~\mathrm dx$$
The summand can be managed with some long division and the digamma function, reducing this down to
$$S=\int_0^1x(x-1)\sum_{n=1}^\infty\frac{(-1)^n}{x+n}~\mathrm dx\\=\frac12\int_0^1x(1-x)\left[\psi^{(0)}\left(\frac{x+2}2\right)-\psi^{(0)}\left(\frac{x+1}2\right)\right]~\mathrm dx$$
Substituting $x\mapsto1-x$ in the second digamma function and using the reflection formula gives us
$$S=\frac12\int_0^1x(1-x)\left[\frac2x-\pi\cot\left(\frac{\pi x}2\right)\right]~\mathrm dx$$
which reduces to a few polylogarithms and the rest is plugging and chugging.
