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T. Amdeberhan
T. Amdeberhan
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Start withby 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}\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} Thereforewhere 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.

Start with 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} 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.

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.

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T. Amdeberhan
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

Start with 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} 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.