any solution?

any ideas?

Bump. This one is quite difficult. seems that the limit for $n \to\infty$ is $\frac{2}{3}$, and it is between zero and one.

I find a method that works for $n\geq 62$. Denote $a_k:=(\prod_{m=1}^{k} \frac{1}{1+\frac{m}{n}})(1+\sum_{m=1}^{k}\frac{m}{n})$ Notice for $k\geq 2$ we have $a_{k-1}-a_k= (\prod_{m=1}^{k} \frac{1}{1+\frac{m}{n}})\frac{k^2(k-1)}{2n^2}.$ So, whenever $k\geq 8$, we have $(\prod_{m=1}^{k} \frac{1}{1+\frac{m}{n}})-\prod_{m=1}^{k} (1-\frac{m}{n})= (\prod_{m=1}^{k} \frac{1}{1+\frac{m}{n}})(1-\prod_{m=1}^{k}(1-\frac{m^2}{n^2}))$ and with Bernoulli $\leq (\prod_{m=1}^{k} \frac{1}{1+\frac{m}{n}})\sum_{m=1}^{k}\frac{m^2}{n^2} = (\prod_{m=1}^{k} \frac{1}{1+\frac{m}{n}}) \frac{k(k+1)(k+0.5)}{3n^2}\leq \frac{51}{56} (\prod_{m=1}^{k} \frac{1}{1+\frac{m}{n}})\frac{k^2(k-1)}{2n^2}=\frac{51}{56}(a_{k-1}-a_k).$ Summing up from k=8 to n-1 we know LHS is less than 51/56. ($a_7<1$) However, for k=1 to 7 we can show with the help of Bernoulli that LHS is less than $\frac{k(k+1)(k+0.5)}{3n^2}$, summing up as $\frac{7*8*8*9}{12n^2}$, which is smaller than 5/56 if $n\geq 62$. So we are sure that the answer is zero for $n\geq 62$.

MazeaLarius wrote: Given an integer $n \geq 2$ determine the integral part of the number $ \sum_{k=1}^{n-1} \frac {1} {({1+\frac{1} {n}}) \dots ({1+\frac {k} {n})}}$ - $\sum_{k=1}^{n-1} (1-\frac {1} {n}) \dots(1-\frac{k}{n})$ who is the author?