We write
$$ S := \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{n+1} $$
for the sum to be computed.
1st Solution. We have
\begin{align*}
S
= \sum_{n=1}^{\infty} \frac{1}{n+1} \sum_{k=2}^{\infty} \frac{1}{k^{2n}}
= \sum_{k=2}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n+1} \frac{1}{k^{2n}}
= \sum_{k=2}^{\infty} \left( - k^2 \log \left( 1 - \frac{1}{k^2} \right) - 1 \right).
\end{align*}
In order to compute this, we write $S_K$ for the partial sums of the last step. Then
\begin{align*}
S_K
&= -K + 1 + \sum_{k=2}^{K} k^2 \log \left( \frac{k^2}{(k+1)(k-1)} \right) \\
&= -K + 1 + \sum_{k=2}^{K} 2 k^2 \log k - \sum_{k=3}^{K+1} (k-1)^2 \log k - \sum_{k=1}^{K-1} (k+1)^2 \log k \\
&= -K + 1 + \log 2 - K^2 \log(K+1) + (K+1)^2 \log K \\
&\quad + \sum_{k=2}^{K} (2 k^2 - (k-1)^2 - (k+1)^2 ) \log k \\
&= -K + 1 + \log 2 - K^2 \log\left(1 + K^{-1}\right) + (2K+1)\log K - 2 \log (K!).
\end{align*}
Now by the Stirling's approximation and the Taylor series of $\log(1+x)$,
$$ 2\log (K!) = \left(2K + 1\right) \log K - 2 K + \log(2\pi) + \mathcal{O}(K^{-1}) $$
and
$$ K^2 \log\left(1 + K^{-1}\right) = K - \frac{1}{2} + \mathcal{O}(K^{-1}) $$
as $K \to \infty$. Plugging this back to $S_K$, we get
$$ S_K = \frac{3}{2} - \log \pi + \mathcal{O}(K^{-1}) $$
and the desired identity follows by letting $K\to\infty$.
2nd Solution. We begin by noting that the Taylor expansion of the digamma function
\begin{align*}
\psi(1+z)
&= -\gamma + \sum_{k=1}^{\infty} (-1)^{k-1} \zeta(k+1) z^{k} \\
&= -\gamma + \zeta(2) z - \zeta(3) z^2 + \zeta(4) z^3 - \dots,
\end{align*}
holds for $|z| < 1$. Then by the Abel's Theorem,
\begin{align*}
S
&= \int_{0}^{1} \sum_{n=1}^{\infty} 2 (\zeta(2n)-1) x^{2n+1} \, \mathrm{d}x \\
&= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(1-x) - \frac{2x}{1-x^2} \right) \, \mathrm{d}x \\
&= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(2-x) + \frac{1}{1+x} \right) \, \mathrm{d}x, \tag{1}
\end{align*}
where the identity
$$ \psi(1+z) = \psi(z) + \frac{1}{z} \tag{2} $$
is used in the last step. Then by using the substitution $x\mapsto 1-x$, we get
$$ \int_{0}^{1} x^2 \psi(2-x) \, \mathrm{d}x
= \int_{0}^{1} (1-x)^2 \psi(1+x) \, \mathrm{d}x. $$
Plugging this back to $\text{(1)}$ and performing integration by parts,
\begin{align*}
S
&= \int_{0}^{1} (2x-1) \psi(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x \\
&= -2 \int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x.
\end{align*}
Now the integrals in the last step can be computed as
$$ \int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x = -1 + \frac{1}{2}\log(2\pi)
\qquad \text{and} \qquad
\int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x = -\frac{1}{2} + \log 2. $$
For instance, the first integral can be computed by writing $\log\Gamma(x+1) = \log\Gamma(x) + \log x$ and applying the Euler's reflection formula. For a more detail, check this posting. Finally, plugging these back to $S$ proves the desired identity.
Although there are no closed forms for both rational zeta series, we can evaluate them with definite integrals.
Set $f(x)$ as
\begin{align*}
f(x)&=\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n-1}x^{n-1}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{1}{k^n}\int_{0}^{x}t^{n-2}\,dt=\int_{0}^{x}\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{t^{n-2}}{k^n}\,dt \\
&=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\sum_{n=2}^{\infty}\left( \frac{t}{k} \right)^{n-2}dt=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\frac{1}{\left(1-\frac{t}{k}\right)}\,dt \\
&=\int_{0}^{x}\left(\frac{1}{t-1}+\sum_{k=1}^{\infty}\frac{1}{k}\frac{1}{(k-t)} \right)dt=\int_{0}^{x}\frac{1}{t-1}\,dt-\int_{0}^{x}\frac{\gamma+\psi(1-t)}{t}\,dt \\
&=\ln(1-x)+\left [ \frac{\ln\Gamma(1-t))}{t} \right ]_{t=0}^{t=x}+\int_{0}^{x}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \\
&=\ln(1-x)+\frac{\ln\Gamma(1-x)}{x}-\gamma+\int_{0}^{x}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \\ \\
f(1)&=\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n-1} \\
&=\lim_{x\rightarrow 1^{-}}\left ( \ln(1-x))+\frac{\ln\Gamma(1-x))}{x} \right )-\gamma+\int_{0}^{1}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \\
&=-\gamma+\int_{0}^{1}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \approx0.7885306 \\
\end{align*}
In the same way, set $f(x)$ as
\begin{align*}
f(x)&=\sum_{n=2}^{\infty}\frac{(-1)^{n}(\zeta(n)-1)}{n-1}x^{n-1}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{(-1)^{n}}{k^n}\int_{0}^{x}t^{n-2}\,dt=\int_{0}^{x}\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{(-t)^{n-2}}{k^n}\,dt \\
&=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\sum_{n=2}^{\infty}\left( -\frac{t}{k} \right)^{n-2}dt=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\frac{1}{\left(1+\frac{t}{k}\right)}\,dt \\
&=\int_{0}^{x}\left(-\frac{1}{t+1}+\sum_{k=1}^{\infty}\frac{1}{k}\frac{1}{(k+t)} \right)dt=-\int_{0}^{x}\frac{1}{t+1}\,dt+\int_{0}^{x}\frac{\gamma+\psi(1+t)}{t}\,dt \\
&=-\ln(1+x)+\left [ \frac{\ln\Gamma(1+t))}{t} \right ]_{t=0}^{t=x}+\int_{0}^{x}\left ( \frac{\ln\Gamma(1+t)}{t^2}+\frac{\gamma}t{} \right )dt \\
&=-\ln(1+x)+\frac{\ln\Gamma(1+x)}{x}+\gamma+\int_{0}^{x}\left ( \frac{\ln\Gamma(1+t)}{t^2}+\frac{\gamma}t{} \right )dt \\ \\
f(1)&=\sum_{n=2}^{\infty}\frac{(-1)^{n}(\zeta(n)-1)}{n-1} \\
&= -\ln(2)+\gamma+\int_{0}^{1}\left ( \frac{\ln\Gamma(1+t)}{t^2}+\frac{\gamma}t{} \right )dt \approx0.5645997 \\
\end{align*}
Best Answer
Concerning the last sum $$S=\sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m}$$ I also do not think that a closed form exist.
However, I think that we could have a good approximation of it making $$S_p=\sum_{m=2}^{p} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m}+\sum_{m=p+1}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m}$$ and, for the second sum, use the asymptotics and get for the summand $$\frac{\pi ^2}{6 m^2}+\frac{\zeta (3)-1}{m^3}+\frac{\pi ^4-45}{90 m^4}+\frac{3\zeta (5)-2}{3m^5}+\frac{2 \pi ^6-945}{1890 m^6}+O\left(\frac{1}{m^7}\right)$$ This makes $S_p$ converging "relatively" fast $$\left( \begin{array}{cc} p & S_p \\ 2 & 1.179744989 \\ 3 & 1.180068403 \\ 4 & 1.180106530 \\ 5 & 1.180113997 \\ 6 & 1.180115992 \\ 7 & 1.180116649 \\ 8 & 1.180116902 \\ 9 & 1.180117011 \\ 10 & 1.180117063 \\ 11 & 1.180117089 \\ 12 & 1.180117103 \\ 13 & 1.180117111 \\ 14 & 1.180117115 \\ 15 & 1.180117118 \\ 16 & 1.180117120 \\ 17 & 1.180117121 \\ 18 & 1.180117122 \\ 19 & 1.180117123 \\ 20 & 1.180117123 \end{array} \right)$$
Computed with illimited precision $$S=1.180117124088625547636095943915451285052438548709796233538676614110$$ which is not recognized by inverse symbolic calculators.