Let $H_n=\sum_{k=1}^n\frac 1 k$ be the $n$-th harmonic number with $H_0=0.$

Question: Is the following true? $$\det\left(H_{i+j}\right)_{i,j=0}^n=(-1)^n \frac{2H_{n}}{n! \prod_{j=1}^n \binom{2j}{j} \binom{2j-1}{j}}.$$

Edit: Comparing with the orthogonal polynomials whose moments are the numbers $\frac{1}{n+1}$ it suffices to show the following identity: $$\sum_{j=0}^n (-1)^j\frac{\binom n j \binom{n+j} j}{\binom{2n} n} H_j \prod_{j=0}^{n-1}\frac{(j!)^3}{(n+j)!} = (-1)^n \frac{2H_n}{n! \prod_{j=1}^n \binom{2j}{j} \binom{2j-1}{j}}.$$

  • $\begingroup$ Have you check it for small $n$? $\endgroup$ – Fedor Petrov Feb 14 '17 at 12:36
  • $\begingroup$ yes, I have checked it for n<50 $\endgroup$ – Johann Cigler Feb 14 '17 at 12:43
  • $\begingroup$ Related: conjecture 3.9 in arxiv.org/abs/1308.2900 $\endgroup$ – Steve Huntsman Feb 14 '17 at 21:11
  • $\begingroup$ See also Krattenthaler's determinant papers, i.e. section 2.7 of arxiv.org/abs/math/9902004 and section 5.4 of arxiv.org/abs/math/0503507 $\endgroup$ – Steve Huntsman Feb 14 '17 at 21:25
  • 1
    $\begingroup$ @JohannCigler it looks that almost all factorials may be cancelled and we may rewrite your identity as $\sum_{j=0}^n (-1)^j\binom{n}{j}\binom{n+j}{j} H_j= 2(-1)^n H_{n}$ $\endgroup$ – Fedor Petrov Feb 16 '17 at 19:44

I prove your identity $$\sum_{j=0}^n (-1)^j\binom{n}{j}\binom{n+j}{j} H_j= 2(-1)^n H_{n}$$ which you claim to imply the result.

The method is the same as here.

At first, use $(-1)^k\binom{n+k}k=\binom{-n-1}k$. Then $$F(y):=\sum_k (-1)^k\binom{n}k\binom{n+k}ky^k=[x^n] (1+x)^n(1+xy)^{-n-1}.$$ Next, for any polynomial $F(y)=\sum c_ky^k$ we have $$ \sum c_kH_k=\int_0^1 \frac{F(y)-F(1)}{y-1}dy. $$ Integration over $[0,1]$ and taking the coefficient of $x^n$ commute, thus we have to prove $$ [x^n]\int_0^1\frac{(\frac{x+1}{1+xy})^n\cdot \frac1{1+xy}-\frac1{1+x}}{y-1}dy=2(-1)^nH_n. $$ A natural change of variables here is $t=(1+x)/(1+xy)$, we get that our integral equals $$-\frac{1}{1+x}\int_1^{1+x}\frac{1-t^{n+1}}{t(1-t)}dt=\frac{-\log(1+x)+H_n}{1+x}-\sum_{i=1}^n\frac{(1+x)^{i-1}}i.$$ A coefficient of $x^n$ indeed equals $2(-1)^nH_n$.

  • $\begingroup$ Would you please explain how does it allow to calculate the determinant? $\endgroup$ – Fedor Petrov Feb 17 '17 at 10:00

As asked by Fedor Petrov I sketch the missing details.

If $a(n)$ is any sequence with $a(0)=1$, such that all Hankel determinants $M_n=\det\left(a(i+j)\right)_{i,j=0}^n$ are $ \neq 0$, define a linear functional $L$ on the polynomials by $L(x^n)=a(n).$ Let $p_n(x)$ be the uniquely determined monic polynomials which are orthogonal with respect to $L.$ These polynomials are given by $$M_{n-1}p_n(x)= \det\left(r(i,j,x)\right)_{i,j=0}^n$$ with $r(i,j,x)=a(i+j)$ for $j<n$ and $r(i,n,x)=x^i.$

For $a(n)=\frac{1}{n+1}$ the corresponding polynomials are $p_n(x)=\sum_{j=0}^n (-1)^j\frac{\binom{n}{j}\binom{n+j}{j}}{\binom{2n}{n}}x^j.$ In this case we get $M_{n-1}=\prod_{j=0}^{n-1}\frac{(j!)^3}{(n+j)!}$ (This seems to be well known, cf. e.g. this preprint (4.2) for $a=b=q=1.$)

Now $\det\left(H_{i+j}\right)_{i,j=0}^n$ can be reduced by column operations to $\det\left(v(i,j)\right)_{i,j=0}^n$, where $v(i,0)=H_{i}$ and $v(i,j)=\frac{1}{i+j}$ for $j>0$. This is the same as replacing $x^i$ in $r(i,n,x)$ by $H_{i}.$ Therefore we get the above identity.

  • $\begingroup$ But there are many monic polynomials of degree $n$ for which $L(p)=0$. $\endgroup$ – Fedor Petrov Feb 17 '17 at 21:18
  • $\begingroup$ Sorry, the mistake has been corrected. $\endgroup$ – Johann Cigler Feb 18 '17 at 9:44
  • $\begingroup$ You mean that $L$ define a scalar product of polynomials by $(f,g)=L(fg)$? $\endgroup$ – Fedor Petrov Feb 18 '17 at 9:48
  • $\begingroup$ @ Fedor Petrov: Yes. $\endgroup$ – Johann Cigler Feb 18 '17 at 11:20

We propose a proof (somewhat different from Fedor's) for the crucial relation $$\sum_{j=0}^n (-1)^j\binom{n}{j}\binom{n+j}{j} H_j= 2(-1)^n H_{n}.\tag1$$ To this end, define the polynomials $$P_n(x):=\sum_{j=0}^n (-1)^j\binom{n}{j}\binom{n+j}{j}\binom{x+j}j.$$ Zeilberger's algorithm returns the recurrence $$(n+2)^2P_{n+2}(x)+(2n+3)(2x+1)P_{n+1}(x)-(n+1)^2P_n(x)=0.\tag2$$ Using the fact that $[x]\binom{x+j}j=H_j,\, [x]x\binom{x+j}j=1,\, P_{n+1}(0)=(-1)^{n+1}$ (see Remark below), induction on equation (1) and applied to (2) leads to: $$(n+2)^2[x]P_{n+2}(x)+(2n+3)[2(-1)^{n+1}+2(-1)^{n+1}H_{n+1}]-(n+1)^22(-1)^nH_n=0.$$ A direct simplification shows $$(n+2)^2[x]P_{n+2}(x) =2(n+2)^2(-1)^{n+2}H_{n+2},$$ which completes the induction process and the proof.

Remark. The identity $(-1)^nP_n(0)=\sum_{j=0}^n (-1)^{n-j}\binom{n}{j}\binom{n+j}{j}=1$ is easily provable by the Wilf-Zeilberger methodology. See my answer here as a further illustration.

  • $\begingroup$ This is a very nice proof. Unfortunately I cannot accept two answers. $\endgroup$ – Johann Cigler Feb 19 '17 at 10:04
  • $\begingroup$ Only a minor comment: the middle term of the last formula should be omitted. $\endgroup$ – Johann Cigler Feb 19 '17 at 10:07
  • $\begingroup$ @JohannCigler: I deleted one item, assuming that was what you were pointing to. $\endgroup$ – T. Amdeberhan Feb 19 '17 at 12:25
  • 1
    $\begingroup$ The identity in the remark is Vandermonde-Chu: substitute $(-1)^j\binom{n+j}{j}=\binom{-n-1}j$ and $\binom{n}j=\binom{n}{n-j}$. $\endgroup$ – Fedor Petrov Feb 19 '17 at 13:39
  • $\begingroup$ @FedorPetrov: Yes, that is another way. $\endgroup$ – T. Amdeberhan Feb 19 '17 at 13:46

Identities involving harmonic numbers that are of interest for physicists, Utilitas Mathematica 83 (2010), 291-299, H. Prodinger.

This paper contains the identity (1) as well.

Now starts Johann Cigler's big birthday (in 5 minutes). Hereby, I will send my best regards on the occasion.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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