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

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. Can anyone help?

Update 1 : I crossposted to MO.

Update 2 : 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 $$B_j=\sum_{k_1+k_2=j,k_i\ge 0\in\mathbb Z}\frac{1}{(2k_1+1)(2k_2+1)}.$$

We shall proceed among similar lines as your argument for $N=3$.

We proceed by induction on $N \in \mathbb{Z}^+$ to prove that for all $n \in \mathbb{Z}^+$ and $a \geq N-1$, we have $$\sum_{k_1+k_2+\ldots+k_N=n,\ k_i\ge0\in\mathbb Z}{\frac{1}{\prod_{j=1}^{N}{(ak_j+1)}}}\leq \frac{N}{a+1}$$

When $N=1$, this is trivially true: $$\frac{1}{an+1} \leq \frac{1}{a+1}$$

Suppose that the statement holds for $N=m$. Then for $N=m+1$, $a \geq (m+1)-1=m$ and any $n \in \mathbb{Z}^+$ we have

\begin{align} &\sum_{k_1+k_2+\ldots+k_{m+1}=n,\ k_i\ge0\in\mathbb Z}{\frac{1}{\prod_{j=1}^{m+1}{(ak_j+1)}}} \\ &=\frac{1}{an+(m+1)}\sum_{k_1+k_2+\ldots+k_{m+1}=n,\ k_i\ge0\in\mathbb Z}{\frac{\sum_{j=1}^{m+1}{(ak_j+1)}}{\prod_{j=1}^{m+1}{(ak_j+1)}}} \\ &= \frac{1}{an+(m+1)}\sum_{k_1+k_2+\ldots+k_{m+1}=n,\ k_i\ge0\in\mathbb Z}{\sum_{j=1}^{m+1}{\frac{1}{\prod_{l \not =j}{(ak_l+1)}}}} \\ &=\frac{m+1}{an+m+1}\sum_{k_1+k_2+\ldots+k_{m+1}=n,\ k_i\ge0\in\mathbb Z}{{\frac{1}{\prod_{l=1}^{m}{(ak_l+1)}}}} \\ &=\frac{m+1}{an+m+1}\sum_{j=0}^{n}{\sum_{k_1+k_2+\ldots+k_m=j,\ k_i\ge0\in\mathbb Z}{{\frac{1}{\prod_{l=1}^{m}{(ak_l+1)}}}}} \\ &=\frac{m+1}{an+m+1}(1+\sum_{j=1}^{n}{\sum_{k_1+k_2+\ldots+k_m=j,\ k_i\ge0\in\mathbb Z}{{\frac{1}{\prod_{l=1}^{m}{(ak_l+1)}}}}}) \\ & \leq \frac{m+1}{an+m+1}(1+n(\frac{m}{a+1})) \, \text{by the induction hypothesis since} \, a \geq m-1\\ &=\left(\frac{m+1}{a+1}\right)\left(\frac{a+1+nm}{an+m+1}\right) \\ &=\left(\frac{m+1}{a+1}\right)\left(1-\frac{(a-m)(n-1)}{an+m+1}\right) \\ & \leq \frac{m+1}{a+1} \, \text{since} \, a \geq m \, \text{and} \, n \geq 1 \end{align}

We are thus done by induction. Now putting $a=N-1$ gives $$\sum_{k_1+k_2+\ldots+k_N=n,\ k_i\ge0\in\mathbb Z}{\frac{1}{\prod_{j=1}^{N}{((N-1)k_j+1)}}}\leq \frac{N}{(N-1)+1}=1$$

• Wow! This is what I really wanted to get. You solved my questions in such a short time. Thank you very much. Commented Oct 29, 2013 at 16:49