Identity with Harmonic and Catalan numbers

Question

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.

Answer 1

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 $p$.

Answer 2

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.

Answer 3

This one can also be done using complex variables, using a variant of Lagrange inversion. I get a slightly different formula on the right.

Suppose we seek to find $$[z^n] \log\left(\frac{2}{1+\sqrt{1-4z}}\right).$$ This is given by $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \log\left(\frac{2}{1+\sqrt{1-4z}}\right) \; dz.$$

Now put $1-4z = w^2$ so that $z=1/4(1-w^2)$ and $-2\; dz = w\; dw$ to get $$\frac{1}{2\pi i} \int_{|w-1|=\gamma} \frac{4^{n+1}}{(1-w^2)^{n+1}} \log\left(\frac{2}{1+w}\right) \left(-\frac{1}{2}\right) w \; dw.$$

This is $$-\frac{1}{2} \frac{4^{n+1}}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(1-w)^{n+1}} \frac{1}{(1+w)^{n+1}} \log\left(\frac{1}{1+(w-1)/2}\right) \times w \; dw$$ or $$\frac{1}{2} \frac{(-1)^n\times 4^{n+1}}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(w-1)^{n+1}} \frac{1}{(1+w)^{n+1}} \log\left(\frac{1}{1+(w-1)/2}\right) \times w \; dw.$$

This has two parts, part $A_1$ is $$\frac{1}{2}\frac{(-1)^n\times 4^{n+1}}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(w-1)^{n}} \frac{1}{(1+w)^{n+1}} \log\left(\frac{1}{1+(w-1)/2}\right) \; dw$$ and part $A_2$ is $$\frac{1}{2} \frac{(-1)^n\times 4^{n+1}}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(w-1)^{n+1}} \frac{1}{(1+w)^{n+1}} \log\left(\frac{1}{1+(w-1)/2}\right) \; dw$$

Part $A_1$ is $$\frac{1}{2} \frac{(-1)^n\times 4^{n+1}}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(w-1)^{n}} \frac{1}{(2+(w-1))^{n+1}} \log\left(\frac{1}{1+(w-1)/2}\right) \; dw \\ = \frac{(-1)^n\times 2^{n}}{2\pi i} \int_{|w-1|=\gamma} \frac{1}{(w-1)^{n}} \frac{1}{(1+(w-1)/2)^{n+1}} \log\left(\frac{1}{1+(w-1)/2}\right) \; dw.$$

Extracting coefficients we get $$(-1)^n 2^{n} \sum_{q=0}^{n-2} {q+n\choose n} \frac{(-1)^q}{2^q} \frac{(-1)^{n-1-q}}{2^{n-1-q} \times (n-1-q)}$$ which is $$-2\sum_{q=0}^{n-2} {q+n\choose n} \frac{1}{n-1-q}.$$

Part $A_2$ is $$(-1)^n 2^{n} \sum_{q=0}^{n-1} {q+n\choose n} \frac{(-1)^q}{2^q} \frac{(-1)^{n-q}}{2^{n-q} \times (n-q)}$$ which is $$\sum_{q=0}^{n-1} {q+n\choose n} \frac{1}{n-q}.$$

Re-index $A_1$ to match $A_2$, getting $$-2\sum_{q=1}^{n-1} {q-1+n\choose n} \frac{1}{n-q}.$$

Collecting the two contributions we obtain $$\frac{1}{n} + \sum_{q=1}^{n-1} \left({q+n\choose n} - 2{q-1+n\choose n}\right) \frac{1}{n-q}$$

which is $$\frac{1}{n} + \sum_{q=1}^{n-1} \left(\frac{q+n}{q} {q-1+n\choose n} - 2{q-1+n\choose n}\right) \frac{1}{n-q} \\ = \frac{1}{n} + \sum_{q=1}^{n-1} \frac{n-q}{q} {q-1+n\choose n} \frac{1}{n-q} \\ = \frac{1}{n} + \sum_{q=1}^{n-1} \frac{1}{q} {q-1+n\choose n} \\ = \frac{1}{n} + \sum_{q=1}^{n-1} \frac{(q-1+n)!}{q!\times n!} \\ = \frac{1}{n} + \frac{1}{n} \sum_{q=1}^{n-1} \frac{(q-1+n)!}{q!\times (n-1)!} \\ = \frac{1}{n} + \frac{1}{n} \sum_{q=1}^{n-1} {q-1+n\choose n-1} = \frac{1}{n} \sum_{q=0}^{n-1} {q-1+n\choose n-1}.$$

To evaluate this last sum we use the integral $${n-1+q\choose n-1} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n-1+q}}{z^{n}} \; dz$$ which gives for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z^{n}} \sum_{q=0}^{n-1} (1+z)^q \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z^{n}} \frac{(1+z)^n-1}{1+z-1} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z^{n+1}} ((1+z)^n-1) \; dz.$$ This also has two components, the second is zero and given by $$-\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z^{n+1}} \; dz$$ leaving $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-1}}{z^{n+1}} \; dz$$ which evaluates to $${2n-1\choose n}.$$

We have shown that $$[z^n] \log\left(\frac{2}{1+\sqrt{1-4z}}\right) = \frac{1}{n} {2n-1\choose n}.$$

Answer 4

Addendum Feb 27 2022. It appears from the comments that OP wanted to prove

$$[z^n] \log^2 \frac{2}{1+\sqrt{1-4z}} = {2n\choose n} (H_{2n-1}-H_n) \frac{1}{n}.$$

Using the result from the companion answers

$$[z^n] \log \frac{2}{1+\sqrt{1-4z}} = \frac{1}{n} {2n-1\choose n}$$

the LHS becomes

$$\sum_{k=1}^{n-1} \frac{1}{k} {2k-1\choose k} \frac{1}{n-k} {2n-2k-1\choose n-k}.$$

Using

$$\frac{1}{k} \frac{1}{n-k} = \frac{1}{n} \frac{1}{k} + \frac{1}{n} \frac{1}{n-k}$$

this becomes

$$\frac{2}{n} \sum_{k=1}^{n-1} \frac{1}{n-k} {2k-1\choose k} {2n-2k-1\choose n-k} \\ = \frac{1}{2n} \sum_{k=1}^{n-1} \frac{1}{n-k} {2k\choose k} {2n-2k \choose n-k} \\ = -\frac{1}{2n^2} {2n\choose n} + \frac{1}{2n} [w^n] \log\frac{1}{1-w} \sum_{k\ge 0} w^k {2k\choose k} {2n-2k \choose n-k}.$$

Here we have extended to infinity due to the coefficient extractor in w (note that $\log\frac{1}{1-w} = w+\cdots$) and canceled the value for $k=0$ that was included in the sum. Continuing with the inner sum term

$$[z^n] \frac{1}{\sqrt{1-4wz}} \frac{1}{\sqrt{1-4z}} \\ = [z^n] \frac{1}{\sqrt{(1-4z)^2 - 4z (1-4z) (w-1)}} \\ = [z^n] \frac{1}{1-4z} \frac{1}{\sqrt{1-4z(w-1)/(1-4z)}} \\ = [z^n] \sum_{k=0}^n {2k\choose k} z^k (w-1)^k \frac{1}{(1-4z)^{k+1}}.$$

This is

$$\frac{1}{2n} [w^n] \log\frac{1}{1-w} \sum_{k=0}^n {2k\choose k} (w-1)^k {n\choose k} 4^{n-k}.$$

Recall from MSE 4316307 that with $1\le k\le n$

$$\frac{1}{k} = {n\choose k} [w^n] \log\frac{1}{1-w} (w-1)^{n-k}.$$

Hence we get two pieces, the first is

$$\frac{1}{2n} \sum_{k=0}^{n-1} {2k\choose k} \frac{1}{n-k} 4^{n-k}.$$

and

$$\frac{1}{2n} [w^n] \log\frac{1}{1-w} {2n\choose n} (w-1)^n.$$

We get for the second

$${2n\choose n} \frac{1}{2n} \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \log\frac{1}{1-w} (-1)^n (1-w)^n.$$

We put $w/(1-w) = v$ so that $w=v/(1+v)$ and $dw = 1/(1+v)^2 \; dv$ to get (without the scalar in front)

$$\;\underset{v}{\mathrm{res}}\; \frac{1}{v^{n+1}} (1+v) \log\frac{1}{1-v/(1+v)} (-1)^n \frac{1}{(1+v)^2} \\ = \;\underset{v}{\mathrm{res}}\; \frac{1}{v^{n+1}} (-1)^n \frac{1}{1+v} \log(1+v) = - (-1)^n [v^n] \frac{1}{1+v} \log\frac{1}{1+v} \\= - [v^n] \frac{1}{1-v} \log\frac{1}{1-v}.$$

With the scalar we get

$$- {2n\choose n} \frac{1}{2n} H_n.$$

We have the result if we can show that the first piece is

$${2n\choose n} \left(H_{2n-1} + \frac{1}{2n} - \frac{1}{2} H_n\right) \frac{1}{n} = {2n\choose n} \left(H_{2n} - \frac{1}{2} H_n\right) \frac{1}{n}$$

i.e.

$$F_n = \sum_{k=0}^{n-1} {2k\choose k} \frac{1}{n-k} 4^{n-k} = {2n\choose n} (2H_{2n} - H_n).$$

We have for the LHS

$$4^n [w^n] \log\frac{1}{1-w} \sum_{k=0}^{n-1} {2k\choose k} w^k 4^{-k}.$$

The coefficient extractor enforces the upper limit, we may extend to infinity and we find

$$4^n [w^n] \log\frac{1}{1-w} \frac{1}{\sqrt{1-w}} = [w^n] \log\frac{1}{1-4w} \frac{1}{\sqrt{1-4w}}.$$

Call the OGF $F(w).$ We get

$$F'(w) = \frac{4}{\sqrt{1-4w}^3} + \frac{2}{1-4w} F(w).$$

Extracting the coefficient on $[w^n]$ we get

$$(n+1) F_{n+1} = 4^{n+1} (-1)^n {-3/2\choose n} + 2 \sum_{q=0}^n F_q 4^{n-q} \\ = 4^{n+1} (-1)^n \frac{n+1}{(-1/2)} {-1/2\choose n+1} + 2 \sum_{q=0}^n F_q 4^{n-q} \\ = 2 (n+1) {2n+2\choose n+1} + 2 \sum_{q=0}^n F_q 4^{n-q}$$

which also yields

$$\frac{1}{4} (n+2) F_{n+2} = \frac{1}{2} (n+2) {2n+4\choose n+2} + 2\sum_{q=0}^{n+1} F_q 4^{n-q}.$$

Subtract to get

$$\frac{1}{4} (n+2) F_{n+2} \\ = (n+1) F_{n+1} + \frac{1}{2} (n+2) {2n+4\choose n+2} - 2 (n+1) {2n+2\choose n+1} + \frac{1}{2} F_{n+1}.$$

Introducing $G_n = F_n {2n\choose n}^{-1}$ and dividing by ${2n+2\choose n+1}$ we get

$$\frac{1}{2} (2n+3) G_{n+2} = (n+3/2) G_{n+1} + 1 \quad\text{or}\quad G_{n} = G_{n-1} + \frac{1}{n-1/2}.$$

so that

$$G_n = \sum_{q=1}^n \frac{1}{q-1/2} = 2 \sum_{q=1}^n \frac{1}{2q-1} = 2 H_{2n-1} - H_{n-1} = 2 H_{2n} - H_n.$$

This is the claim (we have $F_0=G_0=0$ from the generating function) and it completes the entire argument.