Prove that the sum is always greater than $1$ Prove that $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+ \frac{1}{n^2} \ge 1$, for all natural numbers $n$.
I tried to use mathematical inducion but failed and I tried to figure out a short formula for the sum but I couldnt find any.

Using some calculus you can see that $$\sum_{k=n}^{n^2} \frac{1}{k} \ge \int_{n}^{n^2} \frac{1}{x} dx = \ln(n^2) - \ln(n) = \ln(n).$$
Certainly this is larger than 1 for $n\ge 3$. For $n=1$ the claim follows automatically. For $n=2$ we check $$\frac12 + \frac13 + \frac14 = \frac{6+4+3}{12} = \frac{13}{12}.$$

Since we know this holds for $n=2$, another approach is to show that $f(n)=\sum_{k=n}^{n^2}1/k$ is an increasing function. Since between $f(n)$ and $f(n+1)$ we lose $1/n$ and gain $$\frac{1}{n^2+1} + \frac{1}{n^2+2} + \cdots + \frac{1}{(n+1)^2}$$ we need to show that the sum of the new terms is larger than $1/n$.
We can get a lower bound on this tail: $$\frac{1}{n^2+1} + \frac{1}{n^2+2} + \cdots + \frac{1}{(n+1)^2} > \frac{2n+1}{(n+1)^2}.$$ If this left term is larger than $1/n$ we win. Using some precalculus type reasoning we can see that $$\frac{2n+1}{(n+1)^2} \ge \frac1n$$ is equivalent to $$n^2 - n - 1 \ge 0.$$
This is satisfied for $n \ge 2$.
