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:

$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt \left(1+\frac1{2^2}+\dotsb+\frac1{10^2}\right)+\frac1{10\cdot11}+\dotsb+\frac1{n(n-1)}\\&lt;1.549768...+\frac1{10}\lt\frac53$$
I suggest

$$\sum_{k=1}^n \frac{1}{k^2} \leqslant 1 + \sum_{k=2}^n \frac{1}{k^2 - \frac14} = 1 + \sum_{k=2}^n \left(\frac{1}{k-\frac12} - \frac{1}{k+\frac12}\right) = 1 + \frac23 - \frac{1}{n+\frac12}.$$
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 $N=10$.