$n$ 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:
$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt \left(1+\frac1{2^2}+\dotsb+\frac1{10^2}\right)+\frac1{10\cdot11}+\dotsb+\frac1{n(n-1)}\\<1.549768...+\frac1{10}\lt\frac53$$
I suggest
$$\sum_{k=1}^n \frac{1}{k^2} \leqslant 1 + \sum_{k=2}^n \frac{1}{k^2 - \frac14} = 1 + \sum_{k=2}^n \left(\frac{1}{k-\frac12} - \frac{1}{k+\frac12}\right) = 1 + \frac23 - \frac{1}{n+\frac12}.$$
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.
I don't know if this is "better," but since $1/x^2$ is strictly decreasing, we have
$$\sum_{n=N+1}^\infty{1\over n^2}\lt\int_N^\infty{1\over x^2}\,dx={1\over N}$$
after which a little systematic trial and error gives
$$\sum_{n=1}^\infty{1\over n^2}\lt 1+{1\over4}+{1\over9}+{1\over16}+{1\over25}+{1\over5}={5989\over3600}\lt{6000\over3600}={5\over3}$$
Added later: It occurs to me that my approach is not inherently different from the OP's "stupid" method. We each basically argue that
$$\sum_{n=1}^\infty{1\over n^2}\lt1+{1\over4}+\cdots+{1\over N^2}+{1\over N}$$
for any $N$. If there's anything stupid in the OP's solution (which there isn't, really), it's just that it's not necessary to go all the way out to $N=10$.
