Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction: 
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove that given inequality holds for $n+1$ case by using the given $n$ case. Ok, so this is what I've got by now:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{(n+1)^2}\overset{?}{>}1$$
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1$$
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
From the $n$ case we know that:
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}$$
So we basically have to prove that:
$$1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}\frac{1}{n}$$
Since
$$n^2+1<n^2+2<\dots<n^2+2n+1<2n^2+n$$
for $n\geq2$, then also:
$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$ 
$$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$ 
.
.
$$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$ 
so then we have:
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$

If you accept a proof without induction, here is one:
\begin{align*}
\frac 1n+\Bigl(\frac 1{n+1}+\dots+ \frac1{n^2}\Bigr)>\frac 1n +(n^2-n)\cdot\frac 1{n^2}=\frac 1n+1-\frac1n=1.
\end{align*}
This computation is valid if $n^2>n$, i.e. if n>1.
