generating functions - Harmonic Numbers series I

Can it be shown that
\begin{align}

\sum_{n=1}^{\infty} \binom{2n}{n} \ \frac{H_{n+1}}{n+1} \ \left(\frac{3}{16}\right)^{n} = \frac{5}{3} + \frac{8}{3} \ \ln 2 - \frac{8}{3} \ \ln 3
\end{align}
where $H_{n}$ is the Harmonic number and defined as
\begin{align}
H_{n} = \sum_{k=1}^{n} \frac{1}{k} = \int_{0}^{1} \frac{1-t^{n}}{1-t} \ dt.
\end{align}

Answer

Recall the formula of Catalan number
$$
\small

C_n=\frac{1}{n+1}\binom{2n}{n}
$$
and its generating function
$$
\small
\sum_{n=1}^\infty C_n x^n=\frac{1-\sqrt{1-4x}}{2x}-1
$$
then
$$
\small

\begin{align}
f(x)
&=\sum_{n=1}^{\infty} \binom{2n}{n} \frac{H_{n+1}}{n+1} x^n\\
&=\sum_{n=1}^\infty C_n H_{n+1} x^n\\
&=\sum_{n=1}^\infty C_n x^n\lim\limits_{\varepsilon\to 0}\int_\varepsilon^{1-\varepsilon}\frac{1-t^{n+1}}{1-t}dt\\
&=\lim\limits_{\varepsilon\to 0}\sum_{n=1}^\infty C_n x^n\int_\varepsilon^{1-\varepsilon}\frac{1-t^{n+1}}{1-t}dt\\
&=\lim\limits_{\varepsilon\to 0}\int_\varepsilon^{1-\varepsilon}\frac{1}{1-t}\sum_{n=1}^\infty C_n x^n(1-t^{n+1})dt\\
&=\lim\limits_{\varepsilon\to 0}\int_\varepsilon^{1-\varepsilon}\frac{1}{1-t}\left(\sum_{n=1}^\infty C_n x^n-\sum_{n=1}^\infty C_n x^nt^{n+1}\right)dt\\
&=\lim\limits_{\varepsilon\to 0}\int_\varepsilon^{1-\varepsilon}\frac{1}{1-t}\left(\left(\frac{1-
\sqrt{1-4x}}{2x}-1\right)-t\left(\frac{1-\sqrt{1-4xt}}{2xt}-1\right)\right)dt\\

&=\lim\limits_{\varepsilon\to 0}\int_\varepsilon^{1-\varepsilon}\left(\frac{1}{1-t}\frac{\sqrt{1-4xt}-\sqrt{1-4x}}{2x}-1\right)dt\\
&=-1+\frac{1}{2x}\lim\limits_{\varepsilon\to 0}\int_\varepsilon^{1-\varepsilon}\frac{\sqrt{1-4xt}-\sqrt{1-4x}}{1-t}dt\\
&=-1+\frac{1}{2x}\lim\limits_{\varepsilon\to 0}\left(\int_\varepsilon^{1-\varepsilon}\frac{\sqrt{1-4xt}}{1-t}dt-\sqrt{1-4x}\int_\varepsilon^{1-\varepsilon}\frac{dt}{1-t}\right) \\
&=-1+\frac{1}{2x}\lim\limits_{\varepsilon\to 0}\left(\int_{\sqrt{1-4x\varepsilon}}^{\sqrt{1-4x(1-\varepsilon)}}\frac{2u^2}{1-4x-u^2}du-\sqrt{1-4x}\ln\frac{1-\varepsilon}{\varepsilon}\right) \\
&=-1+\frac{1}{2x}\lim\limits_{\varepsilon\to 0}\left(2(1-4x)\int_{\sqrt{1-4x\varepsilon}}^{\sqrt{1-4x(1-\varepsilon)}}\frac{1}{1-4x-u^2}-2\int_{\sqrt{1-4x\varepsilon}}^{\sqrt{1-4x(1-\varepsilon)}}du-\sqrt{1-4x}\ln\frac{1-\varepsilon}{\varepsilon}\right) \\
&=-1+\frac{1}{2x}\lim\limits_{\varepsilon\to 0}\left(\sqrt{1-4x}\ln\frac{\sqrt{1-4x}+u}{\sqrt{1-4x}-u}\Biggl|_{\sqrt{1-4x\varepsilon}}^{\sqrt{1-4x(1-\varepsilon)}}-2u\Biggl|_{\sqrt{1-4x\varepsilon}}^{\sqrt{1-4x(1-\varepsilon)}}-\sqrt{1-4x}\ln\frac{1-\varepsilon}{\varepsilon}\right) \\
&=-1+\frac{1}{2x}\left(\sqrt{1-4x}\lim\limits_{\varepsilon\to 0}\left(\ln\frac{\sqrt{1-4x}+u}{\sqrt{1-4x}-u}\Biggl|_{\sqrt{1-4x\varepsilon}}^{\sqrt{1-4x(1-\varepsilon)}}-\ln\frac{1-\varepsilon}{\varepsilon}\right)-2\lim\limits_{\varepsilon\to 0}u\Biggl|_{\sqrt{1-4x\varepsilon}}^{\sqrt{1-4x(1-\varepsilon)}}\right) \\
&=-1+\frac{\sqrt{1-4x}}{2x}\lim\limits_{\varepsilon\to 0}\left(\ln\frac{\sqrt{1-4x}+u}{\sqrt{1-4x}-u}\Biggl|_{\sqrt{1-4x\varepsilon}}^{\sqrt{1-4x(1-\varepsilon)}}-\ln\frac{1-\varepsilon}{\varepsilon}\right)-\frac{1}{x}\lim\limits_{\varepsilon\to 0}\left(\sqrt{1-4x(1-\varepsilon)}-\sqrt{1-4x\varepsilon}\right) \\
&=-1+\frac{\sqrt{1-4x}}{2x}\lim\limits_{\varepsilon\to 0}\left(\ln\frac{\sqrt{1-4x}+u}{\sqrt{1-4x}-u}\Biggl|_{\sqrt{1-4x\varepsilon}}^{\sqrt{1-4x(1-\varepsilon)}}-\ln\frac{1-\varepsilon}{\varepsilon}\right)-\frac{\sqrt{1-4x}-1}{x} \\
&=\frac{1-x-\sqrt{1-4x}}{x}+\frac{\sqrt{1-4x}}{2x}\lim\limits_{\varepsilon\to 0}\left(\ln\frac{\sqrt{1-4x}+\sqrt{1-4x(1-\varepsilon)}}{\sqrt{1-4x}-\sqrt{1-4x(1-\varepsilon)}}-\ln\frac{\sqrt{1-4x}+\sqrt{1-4x\varepsilon}}{\sqrt{1-4x}-\sqrt{1-4x\varepsilon}}-\ln\frac{1-\varepsilon}{\varepsilon}\right) \\

&=\frac{1-x-\sqrt{1-4x}}{x}+\frac{\sqrt{1-4x}}{2x}\lim\limits_{\varepsilon\to 0}\ln\frac{\sqrt{1-4x}+\sqrt{1-4x(1-\varepsilon)}}{\sqrt{1-4x}-\sqrt{1-4x(1-\varepsilon)}}\frac{\sqrt{1-4x}-\sqrt{1-4x\varepsilon}}{\sqrt{1-4x}+\sqrt{1-4x\varepsilon}}\frac{\varepsilon}{1-\varepsilon} \\
&=\frac{1-x-\sqrt{1-4x}}{x}+\frac{\sqrt{1-4x}}{2x}\lim\limits_{\varepsilon\to 0}\ln\frac{(\sqrt{1-4x}+\sqrt{1-4x(1-\varepsilon)})^2}{(\sqrt{1-4x})^2-(\sqrt{1-4x(1-\varepsilon)})^2}\frac{(\sqrt{1-4x})^2-(\sqrt{1-4x\varepsilon})^2}{(\sqrt{1-4x}+\sqrt{1-4x\varepsilon})^2}\frac{\varepsilon}{1-\varepsilon} \\
&=\frac{1-x-\sqrt{1-4x}}{x}+\frac{\sqrt{1-4x}}{2x}\lim\limits_{\varepsilon\to 0}\ln\frac{(\sqrt{1-4x}+\sqrt{1-4x(1-\varepsilon)})^2}{-4x\varepsilon}\frac{4x(\varepsilon-1)}{(\sqrt{1-4x}+\sqrt{1-4x\varepsilon})^2}\frac{\varepsilon}{1-\varepsilon} \\
&=\frac{1-x-\sqrt{1-4x}}{x}+\frac{\sqrt{1-4x}}{2x}\lim\limits_{\varepsilon\to 0}\ln\frac{(\sqrt{1-4x}+\sqrt{1-4x(1-\varepsilon)})^2}{(\sqrt{1-4x}+\sqrt{1-4x\varepsilon})^2} \\
&=\frac{1-x-\sqrt{1-4x}}{x}+\frac{\sqrt{1-4x}}{x}\ln\frac{2\sqrt{1-4x}}{\sqrt{1-4x}+1} \\
\end{align}
$$
After substitution $\small x=3/16$ we get
$$
\small

f\left(\frac{3}{16}\right)=\frac{5}{3}-\frac{4}{3}\ln\frac{9}{4}
$$