Mathematical Reflections 2013, Issue 5 - Problem U277

Problem:
For $n \in \mathbb{N}$, $n \geq 2$, find the greatest integer less than $2(e^{\frac{1}{n+1}}+\ldots+e^{\frac{1}{n+n}})$.

Proposed by Marius Cavachi.

Solution:
We claim that the greatest integer less than $2(e^{\frac{1}{n+1}}+\ldots+e^{\frac{1}{n+n}})$ is $2n+1$. Indeed,
$$2(e^{\frac{1}{n+1}}+\ldots+e^{\frac{1}{n+n}}) \geq 2ne^{\frac{1}{2n}} \geq 2n\left(1+\dfrac{1}{2n}\right)=2n+1,$$
but $$2(e^{\frac{1}{n+1}}+\ldots+e^{\frac{1}{n+n}}) \leq 2ne^{\frac{1}{n+1}} < 2n+2,$$
where the last inequality can be obtained observing that $$\dfrac{1}{n+1}=\int_{1}^{1+\frac{1}{n}} \dfrac{1}{1+\frac{1}{n}} \ dx <\int_{1}^{1+\frac{1}{n}} \dfrac{1}{x} \ dx=\log\left(1+\dfrac{1}{n}\right).$$