<s>[INST] 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.
 [/INST] 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 $p$.
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