<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:
$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$ 
$$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$ 
.
.
$$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$ 
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 is an alternate solution. We want to prove
$$
\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1
$$
Or, equivalently, that
$$
\frac{\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}}{n^2 - n + 1}>\frac{1}{n^2 - n + 1}
$$
where the left-hand side is the arithmetic mean of all the fractions. We know that the arithmetic mean is greater than the harmonic mean (equality only when all terms are equal, which isn't the case since $n \geq 2$):
$$
\frac{\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}}{n^2 - n + 1} > \frac{n^2 - n + 1}{n + (n+1) + \cdots + n^2}\\
= \frac{n^2 - n + 1}{\frac{n + n^2}{2}(n^2 - n + 1)} = \frac{2}{n^2 + n}
$$
and the proof is completed by noting that we have
$$
n^2 - 3n + 2 \geq 0\\
2n^2 - 2n + 2 \geq n^2 + n\\
2(n^2 - n + 1) \geq n^2 + n\\
\frac{2}{n^2 + n} \geq \frac1{n^2 - n + 1}
$$
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