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For any natural number $n$, Prove that $$\displaystyle \prod^{n}_{r=1}\bigg(r+\frac{1}{n}\bigg)\leq 2(n!)$$
Trial Solution: Using $\displaystyle \frac{1}{n}\leq 1,2,3,\cdots n$
$\displaystyle \prod^{n}_{r=1}\bigg(1+\frac{2}{n}\bigg)\leq 2\cdot 4\cdot 6\cdots \cdots 2n$
$$\prod^{n}_{r=1}\bigg(1+\frac{2}{n}\bigg)\leq 2^n\cdot n!$$
Could some help me how to prove my original inequality, Thanks
We can verify directly for $n=1,2$. Suppose $n\geq3$. Then $$ \sum_{r=1}^n\log\left(1+\frac1{rn}\right) \leq\sum_{r=1}^n\frac1{rn} \leq\frac{1+1/2+(n-2)/3}{n} \leq\frac{11}{18}<\log 2. $$ Now exponentiate and multiply by $n!$.
It is "easy" to show it at least for large values of $n$. $$\prod^{n}_{r=1}\bigg(r+\frac{1}{n}\bigg)=\left(\frac{1}{n}+1\right)_n$$ where appears Pochammer symbol.
Using asymptotics, $$\log \left(\left(1+\frac{1}{n}\right)_n\right)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({n}\right)\right)+\frac{\log \left({n}\right)+\gamma +\frac{1}{12}}{n}+O\left(\frac{1}{n^{3/2}}\right)$$ while, using Stirling approximation, $$ \log(2(n!))=n (\log (n)-1)+\left(\log \left(2 \sqrt{2 \pi }\right)+\frac{1}{2} \log \left({n}\right)\right)+\frac{1}{12 n}+O\left(\frac{1}{n^{3/2}}\right)$$ making $$\log \left(\left(1+\frac{1}{n}\right)_n\right)-\log(2(n!))=-\log (2)+\frac{\gamma +\log \left({n}\right)}{n}+O\left(\frac{1}{n^{3/2}}\right)$$ which is negative $\,\,\forall n$.
