Mathematical Reflections 2016, Issue 3 - Problem U373

Problem:
Prove the following inequality holds for all positive integers $n \geq 2$,
$$\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1+\dfrac{1}{1+2+\ldots+n}\right)<3.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Since $1+2+\ldots+j=\dfrac{j(j+1)}{2}$ for any $j=2,\ldots,n$, we have $$\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1+\dfrac{1}{1+2+\ldots+n}\right)=\prod_{j=2}^n \left(1+\dfrac{2}{j(j+1)}\right).$$ Using the AM-GM Inequality, we have
$$\prod_{j=2}^n \left(1+\dfrac{2}{j(j+1)}\right) \leq \left(\dfrac{n-1+2\sum_{j=2}^n \frac{1}{j(j+1)}}{n-1}\right)^{n-1}=\left(1+\dfrac{1-\frac{2}{n+1}}{n-1}\right)^{n-1}=\left(1+\dfrac{1}{n+1}\right)^{n-1}.$$
By the Binomial Theorem, we have $$\left(1+\dfrac{1}{n+1}\right)^{n-1}=\sum_{k=0}^{n-1} {n-1 \choose k} \dfrac{1}{(n+1)^k}=\sum_{k=0}^{n-1} \dfrac{(n-1)!}{(n-1-k)!(n+1)^k}\cdot\dfrac{1}{k!}.$$ Since $\dfrac{(n-1)!}{(n-1-k)!(n+1)^k}<1$ and $\dfrac{1}{k!} \leq \dfrac{1}{2^{k-1}}$ for all $k \geq 2$, then
$$\left(1+\dfrac{1}{n+1}\right)^{n-1} \leq \sum_{k=0}^{n-1} \dfrac{1}{k!} < 1+1+\sum_{k=2}^{n-1} \dfrac{1}{2^{k-1}}<3,$$ which gives the desired conclusion.