Here's a nice little Mathematica routine for evaluating Tito's continued fraction with precision prec:
prec = 10^4;
y = N[4, prec];
c = y; d = 0; k = 1;
u = 1; v = y;
While[True,
c = 1 + u/c; d = 1/(1 + u d);
h = c*d; y *= h;
v += 96 k^2 + 8;
c = v + u/c; d = 1/(v + u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-prec, Break[]];
u += 3 k (k + 1) + 1;
k++];
6/y
where I used the Lentz-Thompson-Barnett method for the evaluation.
For prec = 10^4, the thing evaluates in 120 seconds (via AbsoluteTiming[]), giving a result that agrees with $\zeta(3)$ to 10,000 digits.
One can consider the even part of Tito's CF, which converges at twice the rate of the original:
$$\cfrac{6}{5-\cfrac{u_1}{v_1-\cfrac{u_2}{v_2-\cfrac{u_3}{v_3-\cdots}}}}$$
where
$$\begin{align*}
u_k&=k^6\\
v_k&=(17k^2+17k+5)(2k+1)
\end{align*}$$
Here's Mathematica code corresponding to this CF:
prec = 10^4;
y = N[5, prec];
c = y; d = 0; k = 1;
While[True,
u = k^6;
v = (2 k + 1) ((17 k + 17) k + 5);
c = v - u/c; d = 1/(v - u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-prec, Break[]];
k++];
6/y
For prec = 10^4, the thing evaluates in 70 seconds (via AbsoluteTiming[]). There may be further ways to accelerate the convergence of the CF, but I have yet to look into them.
Added, quite a bit later:
As it turns out, the even part I derived is precisely Apéry's CF for $\zeta(3)$ (thanks Américo!). Conversely put, Tito's CF is an extension of Apéry's CF. Here's how to derive Apéry's CF from Tito's CF (while proving convergence along the way).
We start from an equivalence transformation of Tito's CF. A general equivalence transformation of a CF
$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cfrac{a_3}{b_3+\cdots}}}$$
with some sequence $\mu_k, k>0$ looks like this:
$$b_0+\cfrac{\mu_1 a_1}{\mu_1 b_1+\cfrac{\mu_1 \mu_2 a_2}{\mu_2 b_2+\cfrac{\mu_2 \mu_3 a_3}{\mu_3 b_3+\cdots}}}$$
Now, given a CF
$$b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cdots}}$$
one can transform this into a CF of the form
$$b_0+\cfrac{w_1}{1+\cfrac{w_2}{1+\cdots}}$$
where $w_1=\dfrac{a_1}{b_1}$ and $w_k=\dfrac{a_k}{b_k b_{k-1}}$ for $k > 1$, where we used $\mu_k=\dfrac1{b_k}$. Applying this transformation to Tito's CF yields the CF
$$\cfrac{\frac32}{1+\cfrac{w_2}{1+\cfrac{w_3}{1+\cdots}}}$$
where $w_{2k}=\dfrac{k^3}{4(2k-1)^3}$ and $w_{2k+1}=\dfrac{k^3}{4(2k+1)^3}$. (You can easily demonstrate that this transformed CF and Tito's CF have identical convergents.)
At this point, we find that since the $w_k \leq\dfrac14$, we have convergence of the CF by Worpitzky's theorem.
Now, we move on to extracting the even part of this transformed CF. Recall that if a CF has the sequence of convergents
$$u_0=b_0,u_1=b_0+\cfrac{a_1}{b_1},u_2=b_0+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2}},\dots$$
then the even part is the CF whose convergents are $u_0,u_2,u_4,\dots$ (Analogously, there is the odd part with the sequence of convergents $u_1,u_3,u_5,\dots$)
Now, given a CF of the form
$$b_0+\cfrac{w_1}{1+\cfrac{w_2}{1+\cdots}}$$
its even part is the CF
$$b_0+\cfrac{w_1}{1+w_2-\cfrac{w_2 w_3}{1+w_3+w_4-\cfrac{w_4 w_5}{1+w_5+w_6-\cdots}}}$$
Thus, the even part of the previously transformed CF is given by
$$\cfrac{\frac32}{\frac54-\cfrac{\beta_1}{\delta_1-\cfrac{\beta_2}{\delta_2-\cdots}}}$$
where
$$\begin{align*}
\beta_k&=\frac{k^3}{4(2k-1)^3}\frac{k^3}{4(2k+1)^3}=\frac{k^6}{16(2k-1)^3(2k+1)^3}\\
\delta_k&=1+\frac{k^3}{4(2k+1)^3}+\frac{(k+1)^3}{4(2k+1)^3}=\frac{17k^2+17k+5}{4(2k+1)^2}
\end{align*}$$
We're almost there! We only need to perform another equivalence transformation, which I'll split into two steps to ease understanding. First, the easy one with $\mu_k=4$, which yields the CF
$$\cfrac{6}{5-\cfrac{16\beta_1}{4\delta_1-\cfrac{16\beta_2}{4\delta_2-\cdots}}}$$
The last step is to cancel out the odd integer denominators of the $\beta_k$ and $\delta_k$; to do this, we take $\mu_k=(2k+1)^3$; this finally yields the CF
$$\cfrac{6}{5-\cfrac{u_1}{v_1-\cfrac{u_2}{v_2-\cfrac{u_3}{v_3-\cdots}}}}$$
where
$$\begin{align*}
u_k&=k^6\\
v_k&=(17k^2+17k+5)(2k+1)
\end{align*}$$
and this is Apéry's CF.
For completeness, I present a formula for the odd part of Tito's CF, after some post-processing with a few equivalence transformations:
$$\zeta(3)=\frac32-\cfrac{81}{\lambda_1-\cfrac{\eta_1}{\lambda_2-\cfrac{\eta_2}{\lambda_3-\ddots}}}$$
where
$$\begin{align*}
\eta_k&=4\times(4k^4+8k^3+k^2-3k)^3=4\times10^3,\,4\times126^3,\dots\\
\lambda_k&=8\times(68k^6-45k^4+12k^2-1)=8\times34,\,8\times3679,\dots
\end{align*}$$
The formula is somewhat more complicated, and converges at the same rate as the even part.
The paper gives (second-to-last page, right-hand column)
$$\sum_{n\ge1}\frac{1}{n^5\binom{2n}{n}}=\frac{9\sqrt{3}\pi}{8}\color{Purple}{L\left(4,\left(-3\atop\circ\right)\right)}+\frac{\pi^2\zeta(3)}{9}-\frac{19\zeta(5)}{3}. \tag{1}$$
You seek a way to write the $L$-function in purple as a linear combination of Hurwitz $\zeta$ functions.
More generally, let $\chi$ be a Dirichlet character of modulus (period) $m$, and define a 'Kronecker' delta
$$\delta_m(k)=\begin{cases}1 & k\equiv0\bmod m \\ 0 & k\not\equiv 0\bmod m\end{cases}. \tag{2}$$
Notice then that $\delta_m(a-b)$ is $1$ if and only if $a\equiv b\bmod m$. We can therefore decompose $\chi$ as
$$\chi(n)=\sum_{k=0}^{m-1} \chi(k) \delta_m(n-k). \tag{3}$$
Furthermore, the Hurwitz zeta function at $a/m\in[0,1)$ decomposes as
$$\begin{array}{c l} \zeta\left(s,\frac{a}{m}\right) & =\sum_{n=1}^\infty\frac{1}{(n+a/m)^s} \\
& =m^s\sum_{n=1}^\infty\frac{1}{(mn+a)^s} \\
& =m^s\sum_{n\ge1} \frac{\delta_m(n-a)}{n^s}.\end{array} \tag{4}$$
Therefore, we have
$$\begin{array}{c l} L(s,\chi) & =\sum_{n\ge1}\frac{\chi(n)}{n^s} \\
& =\sum_{n\ge1}\frac{1}{n^s}\sum_{k=0}^{m-1}\chi(k)\delta_m(n-k) \\
& =\sum_{k=0}^{m-1}\chi(k)\sum_{n\ge1}\frac{\delta_m(n-k)}{n^s} \\
& =\frac{1}{m^s}\sum_{k=0}^{m-1}\chi(k)\zeta\left(s,\frac{k}{m}\right). \end{array} \tag{5}$$
This formula is listed on Wikipedia's Hurwitz $\zeta$ and Dirichlet $L$-function articles. In particular,
$$L\left(4,\left(\frac{-3}{\circ}\right)\right)=\frac{\zeta\left(4,\frac{1}{3}\right)-\zeta\left(4,\frac{2}{3}\right)}{81} \tag{6}$$
because $\left(\frac{-3}{1}\right)=1$ and $\left(\frac{-3}{2}\right)=-1$ (and $\chi(0)=0$ for all Dirichlet characters). Also see here.
Best Answer
It seems to have escaped attention that these sums may be evaluated using harmonic summation techniques.
Introduce the sum $$S(x; \alpha, p) = \sum_{n\ge 1} \frac{1}{n^p(e^{\alpha n x}-1)}$$ with $p$ a positive odd integer and $\alpha>1$, so that we seek e.g. $2 S(1; \pi\sqrt{2}, 3)+S(1; 2\pi\sqrt{2}, 3).$
The sum term is harmonic and may be evaluated by inverting its Mellin transform.
Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have $$\lambda_k = \frac{1}{k^p}, \quad \mu_k = k \quad \text{and} \quad g(x) = \frac{1}{e^{\alpha x}-1}.$$ We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \frac{1}{e^{\alpha x}-1} x^{s-1} dx = \int_0^\infty \frac{e^{-\alpha x}}{1-e^{-\alpha x}} x^{s-1} dx \\ = \int_0^\infty \sum_{q\ge 1} e^{-\alpha q x} x^{s-1} dx = \sum_{q\ge 1} \int_0^\infty e^{-\alpha q x} x^{s-1} dx \\= \Gamma(s) \sum_{q\ge 1} \frac{1}{(\alpha q)^s} = \frac{1}{\alpha^s} \Gamma(s) \zeta(s).$$
It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x;\alpha,p)$ is given by
$$Q(s) = \frac{1}{\alpha^s} \Gamma(s) \zeta(s) \zeta(s+p) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{1}{k^p} \frac{1}{k^s} = \zeta(s+p)$$ for $\Re(s) > 1-p.$
The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$ which we evaluate by shifting it to the left for an expansion about zero.
First formula.
We take $$Q(s) = \frac{1}{\pi^s\sqrt{2}^s} \left(2 + \frac{1}{2^s}\right) \Gamma(s) \zeta(s) \zeta(s+3).$$
We shift the Mellin inversion integral to the line $s=-1$, integrating right through the pole at $s=-1$ picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{\pi^3\sqrt{2}}{72 x} \quad\text{and}\quad \mathrm{Res}(Q(s)/x^s; s=0) = -\frac{3}{2}\zeta(3)$$ and $$\frac{1}{2}\mathrm{Res}(Q(s)/x^s; s=-1) = \frac{\pi^3\sqrt{2}x}{36}.$$
This almost concludes the proof of the first formula if we can show that the integral on the line $\Re(s) = -1$ vanishes when $x=1.$ To accomplish this we must show that the integrand is odd on this line.
Put $s = -1 - it$ in the integrand to get $$\pi^{1+it} \sqrt{2}^{1+it} \left(2 + 2^{1+it}\right) \Gamma(-1-it) \zeta(-1-it) \zeta(2-it).$$
Now use the functional equation of the Riemann Zeta function in the following form: $$\zeta(1-s) = \frac{2}{2^s\pi^s} \cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$ to obtain (with $s=-1-it$) $$\pi^{1+it} \sqrt{2}^{1+it} \left(2 + 2^{1+it}\right) \zeta(2+it) 2^{-1-it} \pi^{-1-it} \frac{1}{2\cos\left(\frac{\pi (-1-it)}{2}\right)} \zeta(2-it)$$ which is $$ \sqrt{2}^{1+it} \left(2^{-it} + 1\right) \zeta(2+it) \frac{1}{2\cos\left(\frac{\pi (1+it)}{2}\right)} \zeta(2-it)$$ and finally yields $$-\frac{1}{\sin(\pi i t/2)} \left(\sqrt{2}^{-1-it}+\sqrt{2}^{-1+it}\right) \zeta(2+it)\zeta(2-it).$$
It is now possible to conclude by inspection: the zeta function terms and the powers of the square root are even in $t$ and the sine term is odd, so the whole term is odd and the integral vanishes. (We get exponential decay from the sine term.)
Second formula.
We take $$Q(s) = \frac{1}{\pi^s\sqrt{2}^s} \left(4 - \frac{1}{2^s}\right) \Gamma(s) \zeta(s) \zeta(s+5).$$
We shift the Mellin inversion integral to the line $s=-2$ (no pole on the line this time) picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{\pi^5\sqrt{2}}{540 x} \quad\text{and}\quad \mathrm{Res}(Q(s)/x^s; s=0) = -\frac{3}{2}\zeta(5)$$ and $$\mathrm{Res}(Q(s)/x^s; s=-1) = \frac{\pi^5\sqrt{2}x}{540}.$$
It remains to verify that the integrand on the line $\Re(s)=-2$ is odd when $x=1$. Put $s=-2-it$ in the integrand to get $$\pi^{2+it} \sqrt{2}^{2+it} \left(4 - 2^{2+it}\right) \Gamma(-2-it) \zeta(-2-it) \zeta(3-it).$$ Applying the functional equation once again with $s=-2-it$ we obtain $$\pi^{2+it} \sqrt{2}^{2+it} \left(4 - 2^{2+it}\right) \zeta(3+it) 2^{-2-it} \pi^{-2-it} \frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)} \zeta(3-it)$$ which is $$\sqrt{2}^{2+it} \left(2^{-it} - 1\right) \zeta(3+it) \frac{1}{2\cos\left(\frac{\pi(-2-it)}{2}\right)} \zeta(3-it)$$ which is in turn $$\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it} \right) \frac{1}{2\cos\left(\frac{\pi(2+it)}{2}\right)} \zeta(3+it) \zeta(3-it)$$ which finally yields $$-\left(\sqrt{2}^{2-it} - \sqrt{2}^{2+it} \right) \frac{1}{2\cos(\pi i t /2)} \zeta(3+it) \zeta(3-it)$$
The product of the zeta function terms is even, as is the cosine term. The term in front is odd, so the integrand is odd as claimed. (We get exponential decay from the cosine term.)
Third formula.
We take $$Q(s) = \frac{1}{\pi^s\sqrt{2}^s} \left(8 + \frac{1}{2^s}\right) \Gamma(s) \zeta(s) \zeta(s+7).$$
We shift the Mellin inversion integral to the line $s=-3$, integrating right through the pole at $s=-3$ picking up the following residues:
$$\mathrm{Res}(Q(s)/x^s; s=1) = \frac{17 \pi^7\sqrt{2}}{37800 x} \quad\text{and}\quad \mathrm{Res}(Q(s)/x^s; s=0) = -\frac{9}{2}\zeta(7)$$ and $$\mathrm{Res}(Q(s)/x^s; s=-1) = \frac{\pi^7\sqrt{2}x}{1134} \quad\text{and}\quad \frac{1}{2} \mathrm{Res}(Q(s)/x^s; s=-3) = -\frac{\pi^7\sqrt{2}x^3}{4050}.$$
This almost concludes the proof of this third formula if we can show that the integral on the line $\Re(s) = -3$ vanishes when $x=1.$ To accomplish this we must show once more that the integrand is odd on this line.
Put $s = -3 - it$ in the integrand to get $$\pi^{3+it} \sqrt{2}^{3+it} \left(8 + 2^{3+it}\right) \Gamma(-3-it) \zeta(-3-it) \zeta(4-it).$$
By the functional equation we obtain with $s = -3-it$ $$\pi^{3+it} \sqrt{2}^{3+it} \left(8 + 2^{3+it}\right) \zeta(4+it) 2^{-3-it} \pi^{-3-it} \frac{1}{2\cos\left(\pi(-3-it)/2\right)} \zeta(4-it)$$ which is $$\sqrt{2}^{3+it} \left(2^{-it} + 1\right) \zeta(4+it) \frac{1}{2\cos\left(\pi(3+it)/2\right)} \zeta(4-it)$$ which finally yields $$\frac{1}{2\sin(\pi it/2)} \left(\sqrt{2}^{3-it} + \sqrt{2}^{3+it}\right) \zeta(4+it) \zeta(4-it).$$
This concludes it since the two zeta function terms together are even as is the square root term while the sine term is odd, so their product is odd.
A similar yet not quite the same computation can be found at this MSE link.
Another computation in the same spirit is at this MSE link II.