This post is aimed as a second solution (v2.0) to AMM problem 11832 proposed by Donald Kunth, U.S.A.
My first proof in a previous blog entry: AMM problem 11832 by D. Knuth
(I believe this new line of computation gains more insight into the cog and wheels (working principles) behind the creation of the identity, rather than my initial approach, which was designed for dealing with a specific case).
We begin with the a theorem with generating functions:
The following is an excerpt from K. N. Boyadzheiv's paper:
Theorem: Given a pair of sequence $\displaystyle (a_n)_{n \ge 0}$ and $\displaystyle (b_n)_{n\ge 0}$ that are Binomial Inverses of each other, that is they satisfy the relation:
for each $n \ge 0$, then the corresponding generating functions with central binomial coefficient weight are related as follows:
Proof:
Lemma: For complex number $\alpha$ and $|z|$ in suitable region of convergence, we have:
$$\displaystyle \sum\limits_{n=0}^{\infty} (-1)^n\binom{\alpha}{n}a_nz^n = (1+z)^{\alpha}\sum\limits_{n=0}^{\infty}(-1)^n\binom{\alpha}{n}\left(\frac{z}{1+z}\right)^n\left(\sum\limits_{k=0}^{n}\binom{n}{k}a_k\right)$$
which is a consequence of Euler's series transformation formula.
The special case $\displaystyle \alpha = -\frac{1}{2}$ is what we are after, $\displaystyle \binom{-1/2}{n} = \frac{1}{4^n}\binom{2n}{n}$ and making the substitution $z \mapsto 4z$ we get the desired result. [Q.E.D.]
For further details please refer to the original paper itself.
Back to the problem at hand:
We have the Binomial transformation formula: $\displaystyle \sum\limits_{k=1}^{n}(-1)^{n-k}\binom{n}{k}(H_{2k} - H_k) = (-1)^{n-1}\left(\frac{4^n}{2n\binom{2n}{n}} - \frac{1}{2n}\right)$ which can be easily established. Hence, putting it into the machinery provided by the theorem, we get:
$$g(z) := \sum\limits_{n=1}^{\infty} \binom{2n}{n}\left(\frac{4^n}{2n\binom{2n}{n}} - \frac{1}{2n}\right)z^n = \frac{1}{2}\sum\limits_{n=1}^{\infty} \frac{4^nz^n}{n} - \frac{1}{2} \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n}= -\frac{1}{2}\log (1-4z) + \log \left(\frac{1+\sqrt{1-4z}}{2}\right)$$
and as a consequence:
$$f(z) := \sum\limits_{n=1}^{\infty} (-1)^{n-1}\binom{2n}{n} (H_{2n} - H_n)z^n = \frac{g\left(\frac{z}{1+4z}\right)}{\sqrt{1+4z}} = \frac{1}{\sqrt{1+4z}}\log \left(\frac{1+\sqrt{1+4z}}{2}\right)$$
Making the change $z \mapsto -z$ in the above identity we get:
$\displaystyle \sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_n)z^n = \frac{-1}{\sqrt{1-4z}}\log \left(\frac{1+\sqrt{1-4z}}{2}\right)$
Dividing both sides with $z$ and integration leads to D. Knuth's identity, as we saw in my previous proof:
$$\boxed{\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n-1}- H_n)\frac{z^n}{n} = \log^2 \left(\frac{1+\sqrt{1-4z}}{2}\right) = \log^2 (C(z))}$$
My first proof in a previous blog entry: AMM problem 11832 by D. Knuth
(I believe this new line of computation gains more insight into the cog and wheels (working principles) behind the creation of the identity, rather than my initial approach, which was designed for dealing with a specific case).
We begin with the a theorem with generating functions:
The following is an excerpt from K. N. Boyadzheiv's paper:
Theorem: Given a pair of sequence $\displaystyle (a_n)_{n \ge 0}$ and $\displaystyle (b_n)_{n\ge 0}$ that are Binomial Inverses of each other, that is they satisfy the relation:
$\displaystyle a_n = \sum\limits_{k=0}^{n} (-1)^{n-k}\binom{n}{k} b_k$ and $\displaystyle b_n = \sum\limits_{k=0}^{n} \binom{n}{k} a_k$
for each $n \ge 0$, then the corresponding generating functions with central binomial coefficient weight are related as follows:
If $\displaystyle f_{b}(z) = \sum\limits_{n=0}^{\infty} \binom{2n}{n}b_nz^n$ then, $\displaystyle f_a(z) = \sum\limits_{n=0}^{\infty} \binom{2n}{n}a_nz^n = \frac{f_b\left(\frac{z}{1+4z}\right)}{\sqrt{1+4z}}$
Proof:
Lemma: For complex number $\alpha$ and $|z|$ in suitable region of convergence, we have:
$$\displaystyle \sum\limits_{n=0}^{\infty} (-1)^n\binom{\alpha}{n}a_nz^n = (1+z)^{\alpha}\sum\limits_{n=0}^{\infty}(-1)^n\binom{\alpha}{n}\left(\frac{z}{1+z}\right)^n\left(\sum\limits_{k=0}^{n}\binom{n}{k}a_k\right)$$
which is a consequence of Euler's series transformation formula.
The special case $\displaystyle \alpha = -\frac{1}{2}$ is what we are after, $\displaystyle \binom{-1/2}{n} = \frac{1}{4^n}\binom{2n}{n}$ and making the substitution $z \mapsto 4z$ we get the desired result. [Q.E.D.]
For further details please refer to the original paper itself.
Back to the problem at hand:
We have the Binomial transformation formula: $\displaystyle \sum\limits_{k=1}^{n}(-1)^{n-k}\binom{n}{k}(H_{2k} - H_k) = (-1)^{n-1}\left(\frac{4^n}{2n\binom{2n}{n}} - \frac{1}{2n}\right)$ which can be easily established. Hence, putting it into the machinery provided by the theorem, we get:
$$g(z) := \sum\limits_{n=1}^{\infty} \binom{2n}{n}\left(\frac{4^n}{2n\binom{2n}{n}} - \frac{1}{2n}\right)z^n = \frac{1}{2}\sum\limits_{n=1}^{\infty} \frac{4^nz^n}{n} - \frac{1}{2} \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n}= -\frac{1}{2}\log (1-4z) + \log \left(\frac{1+\sqrt{1-4z}}{2}\right)$$
and as a consequence:
$$f(z) := \sum\limits_{n=1}^{\infty} (-1)^{n-1}\binom{2n}{n} (H_{2n} - H_n)z^n = \frac{g\left(\frac{z}{1+4z}\right)}{\sqrt{1+4z}} = \frac{1}{\sqrt{1+4z}}\log \left(\frac{1+\sqrt{1+4z}}{2}\right)$$
Making the change $z \mapsto -z$ in the above identity we get:
$\displaystyle \sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_n)z^n = \frac{-1}{\sqrt{1-4z}}\log \left(\frac{1+\sqrt{1-4z}}{2}\right)$
Dividing both sides with $z$ and integration leads to D. Knuth's identity, as we saw in my previous proof:
$$\boxed{\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n-1}- H_n)\frac{z^n}{n} = \log^2 \left(\frac{1+\sqrt{1-4z}}{2}\right) = \log^2 (C(z))}$$
