An inequality about the sum of some unit fractions with a property

Question

Question : Is the following true for any $n,N\in\mathbb N$?

$$\sum_{k_1+k_2+\cdots+k_N=n,\ k_i\ge0\in\mathbb Z}\frac1{\prod_{j=1}^{N}\{(N-1)k_j+1\}}\le 1$$

Motivation : I've known the $N=3$ case :

$$\sum_{k_1+k_2+k_3=n,\ k_i\ge0\in\mathbb Z}\frac1{(2k_1+1)(2k_2+1)(2k_3+1)}\le 1$$

I proved this inequality by estimating the left hand side with integral. After proving this, I reached the above expectation by using computer. The above expectation seems true, but I'm facing difficulty. I would like to know how to prove this (if it's true) and any relevant references.

Remark : This question has been asked previously on math.SE without receiving any answers.

Update : I'm going to show the proof for $N=3$ case without using integral. This is because it seems that this idea can be generalized (though I'm facing difficulty).

For any non-negative integer $n$, $$\sum_{k_1+k_2+k_3=n,\ k_i\ge0\in\mathbb Z}\frac1{(2k_1+1)(2k_2+1)(2k_3+1)}\le 1$$

Proof : Let $A_n$ be the left hand side, and suppose that $\sum$ represents $\sum_{k_1+k_2+k_3=n,k_i\ge 0\in\mathbb Z}$. Noting that $(2k_1+1)+(2k_2+1)+(2k_3+1)=2n+3$, we get $$\begin{align}A_n & =\sum\frac{(2k_1+1)+(2k_2+1)+(2k_3+1)}{(2n+3)(2k_1+1)(2k_2+1)(2k_3+1)}\\ & =\frac{1}{2n+3}\sum\left\{\frac{1}{(2k_1+1)(2k_2+1)}+\frac{1}{(2k_2+1)(2k_3+1)}+\frac{1}{(2k_3+1)(2k_1+1)}\right\}\\ & =\frac{3}{2n+3}\sum\frac{1}{(2k_1+1)(2k_2+1)}\\ & =\frac{3}{2n+3}\sum_{j=0}^n\sum_{k_1+k_2=j,k_i\ge 0\in\mathbb Z}\frac{1}{(2k_1+1)(2k_2+1)}\\ & \le \frac{3}{2n+3}\left(1+\frac 23 n\right)=1\end{align}$$ Here, I used $$B_0=1, B_j\le \frac 23\ (j=1,2,\cdots,n)$$ where $$B_j=\sum_{k_1+k_2=j,k_i\ge 0\in\mathbb Z}\frac{1}{(2k_1+1)(2k_2+1)}.$$

Answer 1

This is not an answer, but I want to use displays. What you are asking is whether the coefficients of the taylor expansion of $$\left( \sum_{k=0}^\infty \frac{x^k}{(N-1)k+1} \right)^N$$ are all at most 1. Unfortunately the summation is a hypergeometric function that doesn't have a general simplification. You can easily prove the conjecture for small $n$ and all $N$, or for small $N$ and all $n$ by expanding it. In general you might be able to bound it with a contour integral or something like that. I believe it is true.

Answer 2

This should be a comment rather than an answer, since I only have plausible strategies to suggest. But I'm new here and don't have enough reputation to leave comments.

Plausible Strategy #1: As Brendan says, your sums are coefficients in the Taylor expansion of $$S_N(x) = \left( \sum_{k=0}^\infty \frac{x^k}{(N-1)k+1} \right)^N.$$ I computed some expansions of these series and it looks like, perhaps, the coefficient of $x^n$ ($n\ge 2$) increases monotonically w.r.t. $N$. You could try to prove this and also prove that the limit of each coefficient as $N\to\infty$ is 1. I think I can show that $S_N(x)$ approaches $\dfrac 1{1-x}$ pointwise for $x\in (-1,1)$, which, though not sufficient for what you want to do, is at least encouraging.

Plausible Strategy #2: If we group like terms in your unit fraction sum, we can express it as a sum indexed over partitions of $n$. I'm too lazy to type the general formula but here it is for $n=3$, which ought to be suggestive enough: $$\left(\frac{N(N-1)(N-2)}{3!}\cdot\frac 1{N^3}\right) + \left(\frac{N(N-1)}{1!1!}\cdot\frac 1{(2N-1)\cdot N}\right) + \left(\frac{N}{1!}\cdot\frac 1{3N-2}\right)$$ Here the terms correspond to $\pi=(1,1,1),(2,1),(3)$ respectively. The coefficients are just garden-variety multinomial coefficients. Now if we consider the terms corresponding to partitions with $r\ge 3$ parts, their numerators are $< (N-1)^r$ and their denominators are $> k(N-1)^r$, where $k$ is the leading coefficient. So, all of these terms undershoot their limits as $N\to\infty$.

When $n$ is large, almost all partitions of $n$ have 3 or more parts, so (a) your conjecture seems likely to be true and (b) you may be able to prove it by bounding a relatively small and simple subset of the terms.

Answer 3

I'm posting an answer just to inform that the question has received an answer by Ivan Loh on MSE.

https://math.stackexchange.com/questions/520220/sum-k-1k-2-cdotsk-n-n-k-i-ge0-in-mathbb-z-frac1-prod-j-1n-n-1k