5
$\begingroup$

This question is related to Hankel determinants of harmonic numbers.

Let $f(n)=\sum_{k=1}^n \frac{2^k}{k}$ and $r(n)=\sum_{j=0}^n (-2)^{n-j}\binom{n}{j}\binom{n+j}{j}f(j).$

In order to compute the Hankel determinants $\det\left(f(i+j)\right)_{i,j=0}^n$ I need the following identities:

1) $r(2n)=0$,

2) $r(2n+1)=(-1)^n \frac{(2n+1)!}{((2n+1)!!)^2}2^{2n+2}.$

Are these identities known?

$\endgroup$
1
  • $\begingroup$ I am not sure your RHS for $r(2n+1)$ fits. Can you check with mine shown below? $\endgroup$ Feb 19, 2017 at 18:00

1 Answer 1

7
$\begingroup$

Define the sequence $a_n(x):=\sum_{j=0}^n(-2)^{n-j}\binom{n}j\binom{n+j}jx^j$ so that $r(n)=\int_0^2\frac{a_n(x)-a_n(1)}{x-1}dx$.

We need the following fact which follows from the Vandermonde-Chu identity: for $n\geq1$, \begin{align}\sum_{j=0}^n(-1)^j\binom{n}j\binom{n+j}j\frac1{j+1} &=\sum_{j=0}^n\binom{n}{n-j}\binom{-n-1}j\frac1{j+1} \\ &=\frac1{n+1}\sum_{j=0}^n\binom{n+1}{n-j}\binom{-n-1}j=0. \tag1 \end{align} At present, the implication of (1) is that $$\int_0^2a_n(x)dx=2(-2)^n\sum_{j=0}^n(-1)^j\binom{n}j\binom{n+j}j\frac1{j+1}=0. \tag2$$ Next, we apply Zeilberger's algorithm to generate the two recursive relations \begin{align} (n+2)a_{n+2}(x)-2(x-1)(2n+3)a_{n+1}(x)+4(n+1)a_n(x)&=0, \\ (n+2)a_{n+2}(1)\qquad \qquad \qquad \qquad \qquad \qquad+4(n+1)a_n(1)&=0. \end{align} Subtract the $2^{nd}$ equation from the $1^{st}$, divide through by $x-1$ and integrate $\int_0^2(\cdot)dx$. So, $$(n+2)r(n+2)+4(n+1)r(n)=0\qquad \implies \qquad r(n+2)=-\frac{4(n+1)}{n+2}\,r(n);$$ where (2) has been utilized effectively. Initial conditions are $r(0)=0$ and $r(1)=4$. The case $n$ even is transparent. The case $n$ odd also follows from induction on $n$ and the fact that if we write the RHS of your claim for $r(n)$ as $$t(n):=\frac{(-1)^{\lfloor n/2\rfloor}4^n}{n\binom{n-1}{\lfloor n/2\rfloor}}\chi_{odd}(n)$$ then it is easy to check that $\frac{t(n+2)}{t(n)}=-\frac{4(n+1)}{2n+3}$ when $n$ is odd. The proof is now complete.

Note. Here $\chi_{odd}$ is understood as $\chi_{odd}(odd)=1$ and $\chi_{odd}(even)=0$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.