<s>[INST] Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$  or  $\frac{1}{(n+1)^2}<2-\frac{1}{n^2} $
 [/INST] As noted, induction is a more difficult way to prove this.  Here it is.
Claim:
$$
\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2} < 2-\frac{1}{n}
$$
for $n=2,3,4,\cdots$.  First, when $n=2$ we have
$$
\frac{1}{1}+\frac{1}{4} = \frac{5}{4} < \frac{3}{2} = 2-\frac{1}{2}
$$
which is correct.
Now, suppose $n \ge 2$ and
$$
\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2} < 2-\frac{1}{n}
$$
Then
$$
\frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2}+\frac{1}{(n+1)^2} < 2-\frac{1}{n}+\frac{1}{(n+1)^2}
$$
so it suffices to prove
$$
2-\frac{1}{n} + \frac{1}{(n+1)^2} < 2-\frac{1}{n+1}
$$
This is equivalent to
$$
\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}
$$
which holds iff
$$
n+n(n+1) < (n+1)^2
$$
or
$$
2n+n^2 < 1+2n+n^2
$$
which is true.  This completes the induction.
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