<s>[INST] 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:
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$
 [/INST] Here's a "cleaner solution" that's an improvement of the result,  though it doesn't use Induction.
Strengthened claim:
$$ \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{3n-2} \geq 1.$$
Proof: Since $ \frac{ 1}{ 2n-1  - k} + \frac{1}{2n-1 +k} = \frac{2(2n-1)}{(2n-1 - k)(2n-1+k)} \geq \frac{2(2n-1)}{(2n-1)^2} = \frac{2}{2n-1}$, the claim follows by pairing up the extreme ends.
Corollary: For $n \geq 2$, since $ 3n - 2 \leq n^2$ , hence the original statement is true.
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