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 the Arithmetic-Harmonic Inequality:
$$\frac{n^2-n+1}{n+(n+1)+\cdots+n^2}\leq\frac{1}{n^2-n+1}\left(\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{n^2}\right)$$
Now, the denominator of the left-hand fraction is:
$$\sum_{k=0}^{n^2-n}(n+k)=\sum_{k=1}^{n^2-n+1}(n+k-1)=\sum_{k=1}^{n^2-n+1}(n-1)+\sum_{k=1}^{n^2-n+1}k=\frac{n^4+n}{2}$$
Hence, we have that
$$1\leq\frac{2(n^2-n+1)}{n(n+1)}=\frac{2(n^2-n+1)^2}{n(n+1)(n^2-n+1)}=\frac{2(n^2-n+1)^2}{n^4+n}\leq\frac{1}{n}+\cdots+\frac{1}{n^2}$$
We can recover the first inequality here from the fact that $0\leq (n-1)(n-2)$ for all $n\geq 1$.
