\begin{eqnarray}
\zeta(s,q)
&=&
\sum_{k=0}^\infty\frac1{(k+q)^s}
\\
&=&
\sum_{k=1}^\infty\frac1{(k+q-1)^s}
\\
&=&
\sum_{k=1}^\infty\frac1{k^s}\cdot\frac1{\left(1+\frac{q-1}k\right)^s}
\\
&=&
\sum_{k=1}^\infty\frac1{k^s}\sum_{n=0}^\infty\binom{-s}n\left(\frac{q-1}k\right)^n
\\
&=&
\sum_{n=0}^\infty\binom{-s}n(q-1)^n\sum_{k=1}^\infty\frac1{k^{n+s}}
\\
&=&
\sum_{n=0}^\infty\binom{-s}n(q-1)^n\zeta(n+s)\;.
\end{eqnarray}
In your case, $\binom{-2}n=(-1)^n(n+1)$, so
\begin{eqnarray}
\zeta\left(2,\frac14\right)-\zeta\left(2,\frac34\right)
&=&
\sum_{n=0}^\infty\binom{-2}n\left(\left(\frac14-1\right)^n-\left(\frac34-1\right)^n\right)\zeta(n+2)
\\
&=&
\sum_{n=0}^\infty(n+1)\left(\left(\frac34\right)^n-\left(\frac14\right)^n\right)\zeta(n+2)
\;.
\end{eqnarray}
An infinite series whose sum is $\pi/4G$.
Applying the idea of @reuns in the comment.
Let $\beta(s)=L(s,\chi_4)$ be Dirichlet beta function. Then we have
$$
\sum_{n=1}^{\infty} \frac{\chi_4(n)\phi(n)}{n^{s+1}} = \frac{\beta(s)}{\beta(s+1)}. \ \ \ (1)
$$
Put $s=1$, we obtain the infinite series
$$
\sum_{n=1}^{\infty} \frac{\chi_4(n)\phi(n)}{n^2}=\frac{\beta(1)}{\beta(2)}=\frac{\pi}{4G}.
$$
By definition of $\chi_4$ below, we may also write
$$
\sum_{k=0}^{\infty} \frac{(-1)^k\phi(2k+1)}{(2k+1)^2}=\frac{\pi}{4G}.
$$
Derivation of (1):
$$
\begin{align}\frac{L(s,\chi_4)}{L(s+1,\chi_4)}&= \prod_p \frac{ 1-\frac{\chi_4(p)}{p^{s+1}}}{1-\frac{\chi_4(p)}{p^s}}\\
&=\prod_p \left(1-\frac{\chi_4(p)}{p^{s+1}}\right)\left(1+\frac{\chi_4(p)}{p^s} + \frac{\chi_4^2(p)}{p^{2s}} + \cdots \right)\\
&=\prod_p \left(1+\frac{\chi_4(p)}{p^s}\left(1-\frac1p\right) + \frac{\chi_4^2(p)}{p^{2s}}\left(1-\frac1p\right)+\cdots\right)\\
&=\sum_{n=1}^{\infty} \frac{\chi_4(n) \frac{\phi(n)}n}{n^s} \ \textrm{ if } \ \Re s >1.
\end{align}
$$
This derivation requires the change of order of summation, which is valid under absolute convergence. So, this is okay if $\Re s >1$.
The convergence of (1) at $s=1$.
For $s=1$, the convergence is a bit more subtle. We proceed as follows.
$$
\begin{align}
\sum_{n\leq x} \frac{\chi_4(n)\phi(n)}{n^2} &= \sum_{n\leq x} \frac{\chi_4(n)}n \sum_{d|n} \frac{\mu(d)}d \ \left(\textrm{We have} \ \frac{\phi(n)}n=\sum_{d|n}\frac{\mu(d)}d\right)\\
&=\sum_{d\leq x} \sum_{k\leq \frac xd} \frac{\mu(d)\chi_4(d)\chi_4(k)}{d^2k} \ (\textrm{Here, we use} \ n=dk )\\
&=\sum_{d\leq x} \frac{\mu(d)\chi_4(d)}{d^2} \sum_{k\leq \frac xd} \frac{\chi_4(k)}k\\
&=\sum_{d\leq x}\frac{\mu(d)\chi_4(d)}{d^2} \left( L(1,\chi_4) + O\left(\frac dx\right)\right)\\
&=\frac{L(1,\chi_4)}{L(2,\chi_4)} +O\left(\frac1x\right) + O\left(\sum_{d\leq x} \frac1{dx}\right)\\
&=\frac{L(1,\chi_4)}{L(2,\chi_4)} + O\left(\frac{\log x}x\right) \\
&=\frac{\pi}{4G}+ O\left(\frac{\log x}x\right).
\end{align}
$$
Definitions of $\chi_4, \phi, \mu$.
Here $\chi_4(n)$ (Dirichlet character modulo $4$) and $\phi(n)$ (Euler Totient function), $\mu(n)$ (Mobius function) are defined by
$$
\chi_4(n)=\begin{cases} 0 &\mbox{ if } n \mbox{ is even}\\
(-1)^k &\mbox{ if } n=2k+1\end{cases}
$$
$$
\phi(n)=n \prod_{p|n} \left(1-\frac 1p\right) \mbox{ is the number of $1\leq k\leq n$ coprime to $n$. }
$$
$$
\mu(n)=\begin{cases} 0 &\mbox{ if } n \ \mbox{ is not square-free}\\
(-1)^{\omega(n)} &\mbox{ if } n \ \mbox{is square-free, }\end{cases}
$$
where $\omega(n)$ is the number of distinct prime factors of $n$.
Best Answer
Using an integral representation of the digamma function we obtain \begin{align} \operatorname{\psi} \left(\frac{n+1}{2}\right) - \operatorname{\psi}\left(\frac{n}{2}\right) - \frac{1}{n} &= \int \limits_0^1 \left[\frac{t^{\frac{n}{2}-1} - t^{\frac{n}{2} - \frac{1}{2}}}{1-t} - \frac{1}{2} t^{\frac{n}{2} -1}\right] \mathrm{d} t = \frac{1}{2} \int \limits_0^1 \frac{1-\sqrt{t}}{1+\sqrt{t}} \, t^{\frac{n}{2} - 1} \, \mathrm{d} t \\ &= \int \limits_0^1 \frac{1-u}{1+u} \, u^{n-1} \, \mathrm{d} u \end{align} for $n \in \mathbb{N}$. For $\sigma \in \{-1,1\}$ and $x \in \left(0,\frac{\pi}{4}\right)$ this implies \begin{align} & \phantom{=~}\sum \limits_{n=1}^\infty \sigma^{n-1} \left[\operatorname{\psi} \left(\frac{n+1}{2}\right) - \operatorname{\psi}\left(\frac{n}{2}\right) - \frac{1}{n}\right] \sin(2 n x) = \int \limits_0^1 \frac{1-u}{1+u} \, \operatorname{Im} \left[\mathrm{e}^{2\mathrm{i}x} \sum \limits_{n=1}^\infty (\sigma \mathrm{e}^{2\mathrm{i}x} u)^{n-1} \right] \, \mathrm{d} u \\ &= \int \limits_0^1 \frac{1-u}{1+u} \frac{\sin(2x)}{1 + u^2 - 2 \sigma u \cos(2x)} \, \mathrm{d} u \stackrel{u = \frac{1-v}{1+v}}{=} \frac{1}{2} \sin(2x) \int \limits_0^1 \frac{v}{1 - \frac{1 + \sigma \cos(2x)}{2} (1-v^2)} \, \mathrm{d} v \\ &= \frac{- \log \left(\frac{1 - \sigma \cos(2x)}{2}\right) \sin(2x)}{2[1 + \sigma \cos(2x)]} = \begin{cases} - \tan(x) \log(\sin(x)) , & \sigma = 1 \\ - \cot(x) \log(\cos(x)) , & \sigma = -1 \end{cases} \, . \end{align} Therefore, your monster equals $$\int \limits_0^{\pi/4} \left[-\tan(x) \log(\sin(x)) \cot(x) + \cot(x) \log(\cos(x))\tan(x)\right] \, \mathrm{d} x = \int \limits_{0}^{\pi/4} \log(\cot(x)) \, \mathrm{d} x \, ,$$ which is indeed an integral representation of Catalan's constant $\mathrm{G}$.