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.
First of all, you can simplify you equivalent. Indeed, $W(n) \rightarrow +\infty$ thus $e^{1/W(n)} \sim 1$ and $W(n) \sim \ln(n)$ thus $\frac{ne^{1/W(n)}}{eW(n)} \sim \frac{n}{e\ln(n)}$.
Then, as @Command Master suggested it, you can use the Dobinski formula,
$$
B_n = \frac{1}{e}\sum_{k \geqslant 0} \frac{k^n}{k!},
$$
where the $(B_n)$ are the Bell numbers (see https://en.wikipedia.org/wiki/Bell_number), which verify,
$$
B_n \sim \frac{1}{\sqrt{n}}\left(\frac{n}{W(n)}\right)^{n + 1/2}\exp\left(\frac{n}{W(n)} - n - 1\right).
$$
Therefore,
$$
S_n^{1/n} = (eB_n)^{1/n} \sim \frac{1}{n^{1/(2n)}}\left(\frac{n}{W(n)}\right)^{1 + 1/(2n)}\exp\left(\frac{1}{W(n)} - 1\right) \sim \frac{n}{e\ln(n)}.
$$
Notice that $u_n \sim v_n > 0$ for $n$ large enough implies $u_n^{1/n} \sim v_n^{1/n}$ but the reciprocal is false in general, so having an equivalent on $S_n$ (given by the equivalent of Bell's numbers) is much more powerful than having an equivalent on $S_n^{1/n}$.
Best Answer
Starting from the equality $ \frac{\pi}{2}\sqrt{n} = \int_{0}^{n} \sqrt{\frac{1}{x} - \frac{1}{n}} \, \mathrm{d}x$, we get
\begin{align*} \sqrt{n}\left( S(n) - \frac{\pi}{2} \right) &= \sum_{k=1}^{n} \sqrt{\frac{1}{k} - \frac{1}{n}} - \frac{\pi}{2}\sqrt{n} \\ &= \sum_{k=1}^{n} \sqrt{\frac{1}{k} - \frac{1}{n}} - \int_{0}^{n} \sqrt{\frac{1}{x} - \frac{1}{n}} \, \mathrm{d}x \\ &= \sum_{k=1}^{n} \int_{k-1}^{k} \biggl( \int_{x}^{k} \frac{\partial}{\partial s} \sqrt{\frac{1}{s} - \frac{1}{n}} \, \mathrm{d}s \biggr) \, \mathrm{d}x \\ &= \sum_{k=1}^{n} \int_{k-1}^{k} \biggl( \int_{k-1}^{s} \frac{\partial}{\partial s} \sqrt{\frac{1}{s} - \frac{1}{n}} \, \mathrm{d}x \biggr) \, \mathrm{d}s \\ &= -\frac{1}{2} \int_{0}^{n} \frac{s - \lfloor s \rfloor}{s^{3/2}\sqrt{1 - s/n}} \, \mathrm{d}s \\ &= -\frac{1}{2} \int_{0}^{0.2022 n} \frac{s - \lfloor s \rfloor}{s^{3/2}\sqrt{1 - s/n}} \, \mathrm{d}s + \mathcal{O}(n^{-1/2}). \end{align*}
Then by the dominated convergence theorem, this converges to
$$ -\frac{1}{2} \int_{0}^{\infty} \frac{s - \lfloor s \rfloor}{s^{3/2}} \, \mathrm{d}s = \zeta\left(\frac{1}{2}\right) $$
as $n \to \infty$. (Note that $\zeta(s) = -s \int_{0}^{\infty} \frac{x-\lfloor x \rfloor}{x^{1+s}} \, \mathrm{d}x $ for $0 < \operatorname{Re}(s) < 1$, see the entry 25.2.8 of DLMF.)