Croatia_2018

Let us consider the sum: $$S=\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{3n+1}$$ There are $(3n+1)-(n+1)+1=2n+1$ terms in the sum.

We will compare this sum to the integral: $${\int }_{n+1}^{3n+2}\frac{1}{x}dx=\ln (3n+2)-\ln (n+1)=\ln \left(\frac{3n+2}{n+1}\right)$$ Note that for $k=n+1,\dots ,3n+1$, we have: $$\frac{1}{k}>{\int }_k^{k+1}\frac{1}{x}dx$$ Therefore, $$S>{\int }_{n+1}^{3n+2}\frac{1}{x}dx=\ln \left(\frac{3n+2}{n+1}\right)$$ We want to show that $S>1$, so it suffices to show that $\ln \left(\frac{3n+2}{n+1}\right)>1$ for all positive integers $n$.

Let us check: $$\ln \left(\frac{3n+2}{n+1}\right)>1$$ This is equivalent to: $$\frac{3n+2}{n+1}>e$$ Let us solve for $n$: $$\begin{gathered} 3n+2>e(n+1) \\ 3n+2>en+e \\ 3n-en>e-2 \\ (3-e)n>e-2 \end{gathered}$$ Since $e\approx 2.718$, $3-e\approx 0.282$, so for $n$ sufficiently large, this is true. Let's check for small $n$:

For $n=1$: $$\frac{3\cdot 1+2}{1+1}=\frac{5}{2}=2.5<e$$ For $n=2$: $$\frac{3\cdot 2+2}{2+1}=\frac{8}{3}\approx 2.666<e$$ For $n=3$: $$\frac{3\cdot 3+2}{3+1}=\frac{11}{4}=2.75>e$$ So for $n\ge 3$, $\ln \left(\frac{3n+2}{n+1}\right)>1$, and thus $S>1$.

For $n=1$: $$S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=0.5+\mathrm{0.333...}+0.25+0.2=\mathrm{1.283...}>1$$ For $n=2$: $$S=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\mathrm{0.333...}+0.25+0.2+\mathrm{0.166...}+\mathrm{0.142...}=\mathrm{1.091...}>1$$ So the inequality holds for all positive integers $n$.

Therefore, $$\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{3n+1}>1$$ for all positive integers $n$.