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:
$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$ 
$$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$ 
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$$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$ 
so then we have:
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$

In case you want a different approach, here is one:

The red rectangles represent the sum $S_n=\frac 1n+...+\frac 1{n^2}$. Above it is for $n=2$, but the principle is general. From the figure it follows immediately that
$$
\int_n^{n^2}\frac 1x\ dx=\ln(n)<\frac 1n+\frac1{n+1}+...+\frac1{n^2}
$$
and for $n>e$ this implies $S_n>1$.
