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}

Solutions Collecting From Web of "Harmonic Numbers series I"

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

Note: Please note, that this question is strongly related (in fact nearly the same) as this question. Both questions are (essentially) asking for a generating function of

\begin{align*}
A(t):=\sum_{n\ge1}C_{n}H_{n+1}t^n=\sum_{n\ge1}\binom{2n}{n}\frac{H_{n+1}}{n+1}t^n\tag{1}
\end{align*}

with $C_n=\frac{1}{n+1}\binom{2n}{n}$ the Catalan numbers and $H_n=\sum_{k=1}^{n}\frac{1}{k}$ the Harmonic numbers. This one asks for $A\left(\frac{3}{16}\right)$, the other one asks for $-A\left(-\frac{1}{4}\right)$.

The answer(s) of the related question are based upon a paper from Boyadhziev Series with Central Binomial Coefficients, Catalan Numbers, and Harmonic Numbers ($2012$). This paper presents many interesting identities and in fact, the generating function we are searching for is also deduced in this paper.

Many ideas from this paper are already stated in the answers of the related question. Here, I try to summarise these and add some additional ones which could be useful to answer the current question.

Overview: The main themes in order to find the ordinary generating function (ogf) of $A(t)$ are binomial inverse pairs and an interesting variation of Euler’s series transformation formula. They are used in combination with the ogf for the Catalan Numbers and the ogf for the central binomial coefficients $\binom{2n}{n}$.

Note: You will see this approach is not really straight forward. But the benefit is, that we will derive some interesting generating functions in order to finally reach $A(t)$.

We start with a binomial inverse pair. To show the relationship we multiply exponential generating functions (egfs). Let $A(x)=\sum_{n\ge0}a_{n}\frac{x^n}{n!}$ and $B(x)=\sum_{n\ge0}b_{n}\frac{x^n}{n!}$ egfs with $B(x)=A(x)e^x$. Comparing coefficients gives the following

Binomial inverse pair
\begin{align*}
B(x)&=A(x)e^x&A(x)&=B(x)e^{-x}\\
b_n&=\sum_{k=0}^{n}\binom{n}{k}a_k&a_n&=\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}b_k\tag{2}
\end{align*}

Next, we observe the following identity for Harmonic Numbers:

Identity of Harmonic Numbers $H_n$
\begin{align*}
H_n&=\sum_{k=1}^{n}(-1)^{k+1}\binom{n}{k}\frac{1}{k}
\end{align*}

(See e.g. the related question for a proof). And we see, that $(-1)^{n-1}H_n$ and $\frac{1}{n}$ are a binomial inverse pair:

Inverse pair with Harmonic Numbers $H_n$
\begin{align*}
(-1)^{n-1}H_n=\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{n-k}}{k}&&\frac{1}{n}=\sum_{k=1}^{n}\binom{n}{k}(-1)^{k-1}H_k\tag{3}
\end{align*}

Most important for the following is Euler’s series transformation formula: which transforms sequences $(a_n)_{n\ge0}$ to $\left(\sum_{k=0}^{n}\binom{n}{k}a_k\right)_{n\ge 0}$ with the help of ordinary generating functions.

Euler’s series transformation formula:

Given a function $f(t)=\sum_{n\ge0}a_{n}t^n$ analytical on the unit disk, the following representation is valid:
\begin{align*}
\left(a_n\right)_{n\ge 0}&&\left(b_n\right)_{n\ge 0}&=\left(\sum_{k=0}^{n}\binom{n}{k}a_k\right)_{n\ge 0}\\
f(t)=\sum_{n\ge0}&a_nt^n&g(t)&=\sum_{n\ge 0}b_nt^n\\
\end{align*}
$$\text{with}\qquad \qquad f(t)=\frac{1}{1-t}g\left(\frac{t}{1-t}\right)$$

Boyadhziev develops a variation of Euler’s series transformation formula customized for central binomial coefficients:

Euler’s series transformation formula for central binomial coefficients:
\begin{align*}
\left(a_n\right)_{n\ge 0}&&\left(b_n\right)_{n\ge 0}&=\left(\sum_{k=0}^{n}\binom{n}{k}a_k\right)_{n\ge 0}\\
f(t)=\sum_{n\ge0}&\binom{2n}{n}a_nt^n&g(t)&=\sum_{n\ge 0}\binom{2n}{n}b_nt^n\\
\end{align*}
\begin{align*}
\text{with}\qquad \qquad f(t)=\frac{1}{\sqrt{1+4t}}g\left(\frac{t}{1+4t}\right)\tag{4}
\end{align*}

Now, it’s time to harvest:

The following identity is valid:

\begin{align*}
\sum_{n\ge 0}\binom{2n}{n}(-1)^{n-1}H_nt^n=\frac{1}{\sqrt{1+4t}}\sum_{n\ge 1}\binom{2n}{n}\frac{1}{n}\left(\frac{t}{1+4t}\right)^n\tag{5}
\end{align*}

Take $a_n=(-1)^{n-1}H_n$ and apply the binomial pairs $(-1)^{n-1}H_n$ and $\frac{1}{n}$ of $(3)$ together with $(4)$ to get $(5)$.

This was the first step to reach $A(z)$. The next step uses the

Generating function of the central binomial coefficients

$$\sum_{n\ge 0}\binom{2n}{n}t^n=\frac{1}{\sqrt{1-4t}}$$

Rewriting it as

$$\sum_{n\ge 1}\binom{2n}{n}t^{n-1}=\frac{1}{t\sqrt{1-4t}}-\frac{1}{t}$$

and integrating it using the substitution $1-4t=y^2$, $dt=-\frac{1}{2}y dy$ results in

\begin{align*}
\sum_{n\ge 1}\binom{2n}{n}\frac{t^{n}}{n}=2\ln\frac{2}{1+\sqrt{1-4t}}\tag{6}
\end{align*}

The calculation:
\begin{align*}
\sum_{n\ge 1}\binom{2n}{n}\frac{t^{n}}{n}&=\int\frac{dt}{t\sqrt{1-4t}}-\int\frac{dt}{t}+C\\
&=-2\int\frac{dy}{1-y^2}-\ln t + C\\
&=\ln{(1-y)}-\ln(1+y)-\ln t + C\\
&=\ln\frac{1-\sqrt{1-4t}}{1+\sqrt{1+4t}}-\ln t + C\\
&=\ln\frac{(1-\sqrt{1-4t})^2}{4t^2}+C\\
&=2\ln\frac{2}{1+\sqrt{1-4t}}
\end{align*}
by observing that $C=0$ when setting $t=0$.

Now, applying the RHS of $(6)$ in $(5)$ we get

\begin{align*}
\sum_{n\ge 0}\binom{2n}{n}(-1)^{n-1}H_nt^n=\frac{2}{\sqrt{1+4t}}\ln\frac{2\sqrt{1+4t}}{1+\sqrt{1+4t}}\tag{7}
\end{align*}

and replacing $t$ with $-t$ yields.

\begin{align*}
\sum_{n\ge 0}\binom{2n}{n}H_nt^n=\frac{2}{\sqrt{1-4t}}\ln\frac{1+\sqrt{1-4t}}{2\sqrt{1-4t}}\tag{8}
\end{align*}

Note: The identities $(7)$ and $(8)$ are Theorem $1$ in the paper from Boyadhziev (as usual $H_0 := 0$).

Now, integrating $(8)$ and using the same substitution $1-4t=y^2$ as in $(6)$ yields:

\begin{align*}
\sum_{n\ge 0}&\binom{2n}{n}\frac{H_n}{n+1}t^{n+1}\\
&=\sqrt{1-4t}\ln(2\sqrt{1-4t})-(1+\sqrt{1-4t})\ln(1+\sqrt{1-4t})+\ln 2\tag{9}
\end{align*}

Note: The identity $(9)$ is Corollary $2$ in the paper from Boyadzhiev.

We are not yet finished, because we are searching an identity similar to $(8)$ but with $H_{n+1}$ instead of $H_n$.

Now, we observe

\begin{align*}
H_{n+1}&=H_n+\frac{1}{n+1}\\
\frac{H_{n+1}}{n+1}&=\frac{H_n}{n+1}+\frac{1}{(n+1)^2}\tag{10}
\end{align*}

We see, when looking at the RHS of $(10)$, that we need an ogf with $a_n=\binom{2n}{n}\frac{1}{(n+1)^2}$. Therefore we start with the

Generating function for the Catalan Numbers:

\begin{align*}
\sum_{n\ge0}\binom{2n}{n}\frac{t^n}{n+1}=\frac{2}{1+\sqrt{1-4t}}\tag{11}
\end{align*}

Integrating $(11)$ and using the same substitution $1-4t=y^2$ as in $(6)$ yields

\begin{align*}
\sum_{n\ge0}\binom{2n}{n}\frac{t^{n+1}}{(n+1)^2}=\ln(1+\sqrt{1-4t})-\sqrt{1-4t}+1-\sqrt{2}\tag{12}
\end{align*}

Now, it’s really time to harvest: Combining according to $(10)$ the ogfs of $(9)$ and $(12)$ we see:

The following identities are valid:

\begin{align*}
\sum_{n\ge0}\binom{2n}{n}\frac{H_{n+1}}{n+1}t^{n+1}
&=1+\sqrt{1-4t}\left(\ln\frac{2\sqrt{1-4t}}{1+\sqrt{1-4t}}-1\right)\\
\text{resp.}&\tag{13}\\
A(t)=\sum_{n\ge1}\binom{2n}{n}\frac{H_{n+1}}{n+1}t^{n}
&=\frac{1-t}{t}+\frac{\sqrt{1-4t}}{t}\left(\ln\frac{2\sqrt{1-4t}}{1+\sqrt{1-4t}}-1\right)\\
\end{align*}

Note: The identities $(13)$ are stated in section $3$ in the paper from Boyadzhiev.

Now, inserting $t=\frac{3}{16}$ in $A(t)$ gives the answer to this question:
\begin{align*}
\sum_{n\ge1}\binom{2n}{n}\frac{H_{n+1}}{n+1}\left(\frac{3}{16}\right)^n
&=\frac{16}{3}\left(1-\frac{3}{16}+\frac{1}{2}\left(\ln\frac{2}{3}-1\right)\right)\\
&=\frac{5}{3}+\frac{8}{3}\ln\frac{2}{3}
\end{align*}

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$\ds{\sum_{n = 1}^{\infty}{2n \choose n}\,
{H_{n + 1} \over n + 1}\pars{3 \over 16}^{n}
={5 \over 3} + {8 \over 3}\,\ln\pars{2} – {8 \over 3}\,\ln\pars{3}
\approx {\tt 0.5854}:\ {\large ?}}$.

\begin{align}
\mbox{Note that}\quad
H_{n + 1}&=\int_{0}^{1}{1 – t^{n + 1} \over 1 – t}\,\dd t
=-\pars{n + 1}\int_{0}^{1}\ln\pars{1 – t}t^{n}\,\dd t
\end{align}

With $\ds{\mu \equiv {3 \over 16}}$:
\begin{align}&\color{#c00000}{\sum_{n = 1}^{\infty}{2n \choose n}\,
{H_{n + 1} \over n + 1}\pars{3 \over 16}^{n}}
=-1 -\int_{0}^{1}\ln\pars{1 – t}\color{#00f}{\sum_{n = 0}^{\infty}%
{2n \choose n}\,\pars{\mu t}^{n}}\,\dd t\tag{1}
\end{align}

Lets evaluate the “$\ds{\color{#00f}{\mbox{blue sum}}}$”:
\begin{align}&\color{#00f}{\sum_{n = 0}^{\infty}{2n \choose n}\,\pars{\mu t}^{n}}
=\sum_{n = 0}^{\infty}\pars{\mu t}^{n}\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{2n} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z}\sum_{n = 0}^{\infty}\bracks{%
\pars{1 + z}^{2}\mu t \over z}^{n}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{1 \over z}
{1 \over 1 – \pars{1 + z}^{2}\mu t/z}\,{\dd z \over 2\pi\ic}
\\[3mm]&=-\,{1 \over \mu t}\oint_{\verts{z}\ =\ 1}
{1 \over z^{2} + \bracks{2 – 1/\pars{\mu t}}z + 1}\,{\dd z \over 2\pi\ic}
\end{align}
The integrand has one pole inside the contour
$\ds{\pars{~p \equiv {1 – 2\mu t – \root{1 – 4\mu t} \over 2\mu t}~}}$ such that:
\begin{align}
&\color{#00f}{\sum_{n = 1}^{\infty}{2n \choose n}\,\pars{\mu t}^{n}}
=-\,{1 \over \mu t}\,{1 \over 2p + 2 – 1/\pars{\mu t}}
=-\,{1 \over 2p\mu t + 2\mu t – 1}
={1 \over \root{1 – 4\mu t}}
\end{align}

Replacing in $\pars{1}$:
\begin{align}&\!\!\!\!\!\!\!\!\!\!
\color{#c00000}{\sum_{n = 1}^{\infty}{2n \choose n}\,
{H_{n + 1} \over n + 1}\ \overbrace{\pars{3 \over 16}^{n}}
^{\ds{\color{#000}{\mu^{n}}}}}\ =\
-1 – \int_{0}^{1}{\ln\pars{1 – t} \over \root{1 – 4\mu t}}\,\dd t
=-1 – \int_{0}^{1}{\ln\pars{1 – t} \over \root{1 – 3t/4}}\,\dd t\tag{2}
\end{align}
The integral is easily evaluated with the change of variables
$\ds{x \equiv \root{1 – 3t/4}}$:
\begin{align}&\int_{0}^{1}{\ln\pars{1 – t} \over \root{1 – 3t/4}}\,\dd t
=-\,{8 \over 3}\int_{1}^{1/2}\bracks{\ln\pars{4 \over 3} + \ln\pars{x + \half}
+ \ln\pars{x – \half}}\,\dd x
\\[3mm]&={8 \over 3}\bracks{\ln\pars{3 \over 2} – 1}
\end{align}

By replacing this result in $\pars{2}$, we find:
$$\color{#66f}{\large%
\sum_{n = 1}^{\infty}{2n \choose n}\,{H_{n + 1} \over n + 1}\pars{3 \over 16}^{n}
={5 \over 3} + {8 \over 3}\,\ln\pars{2} – {8 \over 3}\,\ln\pars{3}}
\approx {\tt 0.5854}
$$