Identity with Harmonic and Catalan numbers Can anyone help me with this.
Prove that 
$$2\log \left(\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^n}{n+1}\right)=\sum_{n=1}^{\infty}\binom{2n}{n}\left(H_{2n-1}-H_n\right)\frac{x^n}{n}$$
Where $H_n=\sum_{k=1}^{n}\frac{1}{k}$.
The left side is equal to $$2\log(C(x))=2\log\left(\frac{2}{1+\sqrt{1-4x}}\right)$$ where $C(x)=\sum_{n=0}^{\infty}C_n x^n$ is the generating function of the Catalan numbers.
 A: Note: Please note that according to the calculation below the factor $(H_{2n-1}-H_n)$ in OPs RHS should be removed.

The following identity is valid
\begin{align*}
\sum_{n=1}^{\infty}\binom{2n}{n}\frac{x^n}{n}=2\log\left(\frac{2}{1+\sqrt{1-4x}}\right)
\end{align*}

The initial point for all calculations is the generating function of the central binomial coefficient
\begin{align*}
\sum_{n= 0}^{\infty}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}\tag{1}
\end{align*}
convergent for  $|x|<\frac{1}{4}$.

Catalan Numbers:
We can integrate (1) from $0$ to $x$ and obtain
  \begin{align*}
\int_{0}^x\sum_{n= 0}^{\infty}\binom{2n}{n}t^ndt&=\int_{0}^x\frac{dt}{\sqrt{1-4t}}\\
&=\frac{1}{4}\int_{1-4x}^{1}\frac{du}{\sqrt{u}}\tag{2}\\
&=\left.\frac{1}{2}\sqrt{u}\right|_{1-4x}^{1}\\
&=\frac{1}{2}\left(1-\sqrt{1-4x}\right)
\end{align*}
Comment:
  
  
*
  
*In (2) we substitute $u=1-4t$ and $du=-4dt$
  
  
  Now dividing LHS and RHS of (2) by $x$, we get the generating function of the Catalan numbers
  \begin{align*}
\sum_{n= 0}^{\infty}\binom{2n}{n}\frac{x^n}{n+1}&=\frac{1}{2x}\left(1-\sqrt{1-4x}\right)\\
&=\frac{2}{1+\sqrt{1-4x}}
\end{align*}

$$$$

OP's identity: Just another variation based upon (1)
We again perform integration but we put the first term of the series to the right side, divide both sides by $t$ and we observe:
  \begin{align*}
\int_{0}^x\sum_{n=1}^{\infty}\binom{2n}{n}t^{n-1}dt
&=\int_{0}^x\left(\frac{1}{t\sqrt{1-4t}}-\frac{1}{t}\right)dt\\
&=\int_{0}^{x}\frac{1-\sqrt{1-4t}}{t\sqrt{1-4t}}dt\tag{3}\\
&=\int_{1-4x}^{1}\frac{1-\sqrt{u}}{(1-u)\sqrt{u}}du\\
&=\int_{1-4x}^{1}\frac{du}{(1+\sqrt{u})\sqrt{u}}\\
&=2\int_{\sqrt{1-4x}}^{1}\frac{ds}{1+s}\tag{4}\\
&=2\left.\log(1+s)\right|_{\sqrt{1-4x}}^{1}\\
&=2\left(\log 2 - \log \left(1+\sqrt{1-4x}\right)\right)\\
&=2\log\frac{2}{1+\sqrt{1-4x}}
\end{align*}
Comment:
  
  
*
  
*In (3) we substitute $u=1-4t$ and $du=-4dt$
  
*In (4) we substitute $s=\sqrt{u}$ and $ds=\frac{1}{2\sqrt{u}}du$
We obtain from LHS and RHS of (4) the identity
  \begin{align*}
\sum_{n=1}^{\infty}\binom{2n}{n}\frac{x^n}{n}=2\log\left(\frac{2}{1+\sqrt{1-4x}}\right)
\end{align*}
and the claim follows.

Note: OPs identity can be found in Interesting Series involving the Central Binomial Coefficient by D.H. Lehmer. It is also stated in Series with Central Binomial Coefficients, Catalan Numbers, and Harmonic Numbers by K.N. Boyadzhiev with a reference to Lehmer's paper. Both papers contain interesting related identities.
A: We start with the key identity:
$$\sum\limits_{j=1}^{n-k} \frac{(-1)^{j-1}}{j}\binom{n}{k+j} = \binom{n}{k}(H_n - H_k)$$
which can be proved elementarily by induction on $n$ or otherwise.
Thus we have for our special case: $\displaystyle \binom{2n}{n}(H_{2n} - H_n) = \sum\limits_{j=1}^{n} (-1)^{j-1}\frac{1}{j}\binom{2n}{n+j}$
The series then decomposes to two parts: 
$$\displaystyle \sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n-1}- H_n)\frac{z^n}{n} = \underbrace{\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n}- H_n)\frac{z^n}{n}}_{=A} - \frac{1}{2}\underbrace{\sum\limits_{n=1}^{\infty} \binom{2n}{n} \frac{z^n}{n^2}}_{ = B}$$
Evaluation of $A$:
$$\begin{align}\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n}- H_n)\frac{z^n}{n} & = \sum\limits_{n=1}^{\infty} \sum\limits_{j=1}^{n} (-1)^{j-1}\frac{1}{j}\binom{2n}{n+j}\frac{z^n}{n} \\ &= \sum\limits_{j=1}^{\infty} \frac{(-1)^{j-1}}{j}\sum\limits_{n=j}^{\infty} \binom{2n}{n+j}\frac{z^n}{n} \\ &= \sum\limits_{j=1}^{\infty} \frac{(-1)^{j-1}z^{j}}{j}\sum\limits_{m=0}^{\infty} \binom{2j+2m}{2j+m}\frac{z^m}{j+m} \\ &= \int_0^z \frac{1}{z}\sum\limits_{j=1}^{\infty} \frac{(-1)^{j-1}z^{j}}{j}\sum\limits_{m=0}^{\infty} \binom{2j+2m}{2j+m} z^{m} \,\mathrm{d}z \tag{*}\\ &= \int_0^z \frac{1}{z}\sum\limits_{j=1}^{\infty} \frac{(-1)^{j-1}z^j}{j}.\frac{4^j}{\sqrt{1-4z}(1+\sqrt{1-4z})^{2j}} \,\mathrm{d}z\\&= \int_0^z \frac{1}{z\sqrt{1-4z}}\log \left(1+\frac{4z}{(1+\sqrt{1-4z})^2}\right) \,\mathrm{d}z \\ &= -\int_0^z \frac{1}{z\sqrt{1-4z}}\log \left(\frac{1+\sqrt{1-4z}}{2}\right) \,\mathrm{d}z\end{align}$$
Consider, $\displaystyle f(z) = \log \left(\frac{1+\sqrt{1-4z}}{2}\right)$, then $\displaystyle f'(z) = \frac{1}{2z} - \frac{1}{2z\sqrt{1-4z}}$
Continuing the computation with the substitution:
$$\begin{align}& = \int_0^z f(z)\left(2f'(z) - \frac{1}{z}\right)\,\mathrm{d}z \\&= \log^2 \left(\frac{1+\sqrt{1-4z}}{2}\right) - \int_0^z \frac{1}{z}\log \left(\frac{1+\sqrt{1-4z}}{2}\right)\,\mathrm{d}z\end{align}$$
Since, 
$$\displaystyle \begin{align} \sum\limits_{n=1}^{\infty} \binom{2n}{n}z^n = \frac{1}{\sqrt{1-4z}} &\implies \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n} = -2\log \left(\frac{1+\sqrt{1-4z}}{2}\right)\\ & \implies \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n^2} = -2\int_0^z\frac{1}{z}\log \left(\frac{1+\sqrt{1-4z}}{2}\right)\,\mathrm{d}z =B \end{align}$$
And also, $\displaystyle C(z) = \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n+1} = \frac{1}{z}\int_0^z \frac{1}{\sqrt{1-4z}}\,\mathrm{d}z = \frac{1-\sqrt{1-4z}}{2z} = \frac{2}{1+\sqrt{1-4z}}$
Hence, we get our final result:
$$\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n-1}- H_n)\frac{z^n}{n} = \log^2 \left(\frac{1+\sqrt{1-4z}}{2}\right) = \log^2 (C(z))$$
In $(*)$ we used the fact that: $\displaystyle \sum\limits_{m=0}^{\infty} \binom{p+2m}{p+m}z^m = \frac{2^p (1+\sqrt{1-4z})^{-p}}{\sqrt{1-4z}}$, for integers $p$.
