Croatia competition problems

Let us consider the sum:
$$S = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{3n+1}$$
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 $k = n+1, \ldots, 3n+1$, 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 $n$:
$$3n+2 > e(n+1) \\ 3n+2 > en + e \\ 3n - en > e - 2 \\ (3-e)n > e-2$$
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 $n \geq 3$, $\ln\left(\frac{3n+2}{n+1}\right) > 1$, and thus $S > 1$.

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

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