<s>[INST] An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$ $n$ is a positive integer, then 
$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53.$$
please don't refer to the famous  $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.
I want to find a better proof. 
My  stupid method:
$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt \left(1+\frac1{2^2}+\dotsb+\frac1{10^2}\right)+\frac1{10\cdot11}+\dotsb+\frac1{n(n-1)}\\<1.549768...+\frac1{10}\lt\frac53$$
 [/INST] We show by induction that if $n\gt 1$ then
$$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2}\lt \frac{5}{3}-\frac{2}{2n+1}.$$
The result is true at $n=2$. Suppose that the result holds at $n=k$. We show it holds at $n=k+1$.
By the induction assumption,
$$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{k^2}+\frac{1}{(k+1)^2}\lt \frac{5}{3}-\frac{2}{2k+1}+\frac{1}{(k+1)^2}.\tag{1}$$
The right-hand side of (1) is equal to
$$\frac{5}{3}-\left(\frac{2}{2k+1}-\frac{1}{(k+1)^2}\right).$$
Now we need to show that $\frac{2}{2k+1}-\frac{1}{(k+1)^2}\gt \frac{1}{2k+3}$ or equivalently that $\frac{4}{(2k+1)(2k+3)}\gt \frac{1}{(k+1)^2}$. So we show that $4(k+1)^2\gt (2k+1)(2k+3)$. This is straightforward. 
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