I am trying to show that
$$\lim\_{n\to\infty}\frac{1}{n}\sum\_{i=1}^{n}\prod\_{j=0}^{i}\frac{kn-j-k}{kn-j}=\frac{k^{k+1}-(k-1)^{k+1}}{(k+1)k^{k}}$$
for all $k\in\mathbb{N}$, $k\geq 4$.


I could verify the statement with Mathematica, but I could not find a self-standing proof. The product is telescopic, but I could only derive lower bounds instead of the exact value of the limit.



Limit of the average of telescopic products
Start by re-expressing the product term
\begin{align}\frac{(kn-k)(kn-k-1)\cdots(kn-k-i)}{(kn)(kn-1)\cdots(kn-i)}
&=\frac{(kn-i-1)\cdots(kn-i-k)}{(kn)\cdots(kn-k+1)}=\frac{\binom{kn-1-i}k}{\binom{kn}k}.
\end{align}
So, the given sum $S\_n(k):=\sum\_{i=1}^n\prod\_{j=0}^i\frac{kn-j-k}{kn-j}$ reads
\begin{align}
S\_n(k)&=\frac1{\binom{kn}k}\sum\_{i=1}^n\binom{kn-1-i}k
=\frac1{\binom{kn}k}\sum\_{j=0}^{n-1} \binom{kn-n-1+j}{k} \\
&=\frac1{\binom{kn}k}\left[\sum\_{\ell=0}^{kn-2} \binom{\ell}k
-\sum\_{\ell=0}^{kn-n-1} \binom{\ell}k\right] \\
&=\frac1{\binom{kn}k}\left[\binom{kn-1}{k+1}-\binom{kn-n}k\right] \\
&=\frac{kn-1}{k+1}\cdot\prod\_{j=1}^k\frac{kn-1-j}{kn-j}-
\frac{kn-n}{k+1}\cdot\prod\_{j=1}^k\frac{kn-n-1-j}{kn-j} \end{align}
where we have utilized the identity $\sum\_{\ell=0}^N\binom{\ell}k=\binom{N+1}{k+1}$. Therefore, going back to the required limit
\begin{align}
\lim\_{n\rightarrow\infty}\frac1nS\_n(k)
&=\lim\_{n\rightarrow\infty}
\frac{kn-1}{(k+1)n}\cdot\prod\_{j=1}^k\frac{kn-1-j}{kn-j}-
\frac{(k-1)n}{(k+1)n}\cdot\prod\_{j=1}^k\frac{(k-1)n-1-j}{kn-j} \\
&=\frac{k}{k+1}-\frac{k-1}{k+1}\cdot\frac{(k-1)^k}{k^k}
=\frac{k^{k+1}-(k-1)^{k+1}}{(k+1)\,k^k}.
\end{align}
Your claim has been verified.


