TITLE: Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$
QUESTION [4 upvotes]: 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} $

REPLY [1 votes]: The sum $\sum_{k=2}^n\frac{1}{k^2}$ is a lower Riemann sum for the function $1/x^2$ in the interval $[1,n]$, so
$$
\sum_{k=2}^n\frac{1}{k^2}\le\int_{1}^n\frac{1}{x^2}\,dx=
\left[-\frac{1}{x}\right]_1^n=1-\frac{1}{n}
$$
Therefore
$$
\sum_{k=1}^n\frac{1}{k^2}\le 1+1-\frac{1}{n}<2
$$
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