How to show that this identity holds?

Question

By comparing the first terms of the Taylor expansion at $0$, it seems that for $|x|<4/27$ the following identity holds: $$\ln\left(\sum_{n=1}^{\infty} \binom{3n}{n}\frac{x^{n-1}}{2n+1}\right)= \sum_{n=1}^{\infty} \binom{3n}{n}\frac{x^n}{n}.$$ I tried by differentiating both sides, but I am completely lost in a terrible mess. I wonder if there is a better strategy to handle it. Any idea?

Answer 1

Let's try to use your idea of looking at derivatives: For $x=0$ both sides result in $0$, so indeed, if the derivatives are the same then we have the equality.

Let $g(x)$ be the left-hand side and $f(x)$ the right-hand side. The derivative of $g(x)$ is $$g'(x)=\frac{1}{\sum_{n=1}^\infty\binom{3n}{n}\frac{x^{n-1}}{2n+1}}\sum_{n=2}^\infty\binom{3n}{n}\frac{(n-1)}{2n+1}x^{n-2}$$ and the derivative of the right-hand side is $$f'(x)=\sum_{n=1}^\infty\binom{3n}{n}x^{n-1}$$

Just to simplify things, let's rewrite all these sums starting at $0$:

$$g'(x)=\frac{1}{\sum_{n=0}^\infty\binom{3n+3}{n+1}\frac{x^n}{2n+3}}\sum_{n=0}^\infty\binom{3n+6}{n+2}\frac{n+1}{2n+5}x^n$$ $$f'(x)=\sum_{n=0}^\infty\binom{3n+3}{n+1}x^n$$

So $g'(x)=f'(x)$ is equivalent to $$\sum_{n=0}^\infty\binom{3n+6}{n+2}\frac{n+1}{2n+5}x^n=\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}\frac{x^n}{2n+3}\right)\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}x^n\right)$$ Now let's use the fact that $(\sum_{n=0}^\infty a_n)(\sum_{n=0}^\infty b_n)=\sum_{n=0}^\infty\left(\sum_{j=0}^n a_jb_{n-j}\right)$, so \begin{align*} \left(\sum_{n=0}^\infty\binom{3n+3}{n+1}\frac{x^n}{2n+3}\right)\left(\sum_{n=0}^\infty\binom{3n+3}{n+1}x^n\right)\hspace{-80pt}&\\ &=\sum_{n=0}^\infty\left(\sum_{j=0}^n\binom{3j+3}{j+1}\frac{x^j}{2j+3}\binom{3n-3j+3}{n-j+1}x^{n-j}\right)\\ &=\sum_{n=0}^\infty\left(\sum_{j=0}^n\frac{1}{2j+3}\binom{3j+3}{j+1}\binom{3n-3j+3}{n-j+1}\right)x^n. \end{align*}

Finally, the problem is equivalent to showing that for all $n$, $$\binom{3n+6}{n+2}\frac{1}{2n+5}=\sum_{j=0}^n\frac{1}{2j+3}\binom{3j+3}{j+1}\binom{3n-3j+3}{n-j+1}$$ For $n=0$ both sides yield $3$, so perhaps the rest will follow by induction.

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EDIT: For $n=1$, the left-hand above side yields 12 but the right-hand side yields 24, so apparently this is not true (if there is no mistake in my arguments).

Answer 2

Suppose we seek to show that

$$\sum_{n\ge 0} {3n+3\choose n+1} \frac{z^n}{2n+3} = \exp\left(\sum_{n\ge 1} {3n\choose n} \frac{z^n}{n}\right).$$

This is the same as showing that

$$q_n = [z^n] f(z) = {3n+3\choose n+1} \frac{1}{2n+3} \quad\text{where}\quad f(z) = \exp\left(\sum_{n\ge 1} {3n\choose n} \frac{z^n}{n}\right).$$

and it can be shown by induction. Basic arithmetic verifies it for $q_0 = 1.$ Differentiating we get

$$f'(z) = \exp\left(\sum_{n\ge 1} {3n\choose n} \frac{z^n}{n}\right) \left(\sum_{n\ge 1} {3n\choose n} z^{n-1} \right) \\ = f(z) \left(\sum_{n\ge 0} {3n+3\choose n+1} z^{n} \right).$$

Extracting coefficients we obtain the recurrence

$$[z^n] f'(z) = (n+1) q_{n+1} = \sum_{k=0}^n q_k {3n+3-3k\choose n+1-k}$$

or

$$q_{n+1} = \frac{1}{n+1} \sum_{k=0}^n q_k {3n+3-3k\choose n+1-k}.$$

This means to verify the claim we have to show that

$${3n+6\choose n+2} \frac{1}{2n+5} = \frac{1}{n+1} \sum_{k=0}^n {3k+3\choose k+1} \frac{1}{2k+3} {3n+3-3k\choose n+1-k}$$

or

$${3n+6\choose n+2} \frac{n+1}{2n+5} = \sum_{k=0}^{n} {3k+3\choose k+1} \frac{1}{2k+3} {3n+3-3k\choose n+1-k}$$

which is

$${3n+6\choose n+2} \frac{n+2}{2n+5} = {3n+6\choose n+1} \\ = \sum_{k=0}^{n+1} {3k+3\choose k+1} \frac{1}{2k+3} {3n+3-3k\choose n+1-k}.$$

What we have here is a straightforward convolution of two ordinary generating functions. With some help from the OEIS we learn that we require the ternary unlabeled tree function $T(z)$ with functional equation

$$T(z) = 1 + z T(z)^3 \quad\text{or}\quad z = \frac{T(z)-1}{T(z)^3}.$$

We then obtain

$$[z^n] T(z) = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z) \; dz.$$

Using the functional equation and letting $w=T(z)$ we have $dz = (3-2w)/w^4 \; dw$ and get for the integral

$$\frac{1}{2\pi i} \int_{|w-1|=\gamma} \frac{w^{3n+3}}{(w-1)^{n+1}} w \frac{3-2w}{w^4} \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\gamma} \frac{3w^{3n}-2w^{3n+1}}{(w-1)^{n+1}} \; dw.$$

Now with $w^q = \sum_{p=0}^q {q\choose p} (w-1)^p$ this will produce

$$3\times {3n\choose n} - 2\times {3n+1\choose n} = \left(3 - 2\frac{3n+1}{2n+1}\right) {3n\choose n} = \frac{1}{2n+1} {3n\choose n}.$$

This generates the left term of the convolution. For the right term consider

$$[z^n] \frac{1}{1-3zT(z)^2} = [z^n] \frac{1}{1-3(T(z)-1)/T(z)} = [z^n] \frac{T(z)}{3-2T(z)}$$

to get

$$\frac{1}{2\pi i} \int_{|w-1|=\gamma} \frac{w^{3n+3}}{(w-1)^{n+1}} \frac{w}{3-2w} \frac{3-2w}{w^4} \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\gamma} \frac{w^{3n}}{(w-1)^{n+1}} \; dw = {3n\choose n}.$$

Recall the convolution that we are working with which now becomes

$$\sum_{k=0}^{n+1} [z^{k+1}] T(z) [z^{n+1-k}] \frac{T(z)}{3-2T(z)} \\ = \sum_{k=0}^{n+1} [z^k] \frac{T(z) - 1}{z} [z^{n+1-k}] \frac{T(z)}{3-2T(z)}.$$

We are left with

$$[z^{n+1}] \frac{1}{z} \frac{T(z)(T(z)-1)}{3-2T(z)} = [z^{n+2}] \frac{T(z)(T(z)-1)}{3-2T(z)}.$$

Using the sáme integral as before this becomes

$$\frac{1}{2\pi i} \int_{|w-1|=\gamma} \frac{w^{3n+9}}{(w-1)^{n+3}} \frac{w(w-1)}{3-2w} \frac{3-2w}{w^4} \; dw \\ = \frac{1}{2\pi i} \int_{|w-1|=\gamma} \frac{w^{3n+6}}{(w-1)^{n+2}} \; dw \\ = {3n+6\choose n+1}.$$

This concludes the argument.

The sequences where this material resides are OEIS A005809 and OEIS A001764.

The structure of the initial equation suggests a counting argument involving sets of labeled directed cycles where each item may have one of three colors and the reader is invited to investigate this further.