If $n$ is a positive integer, Prove that $\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac{2329}{3600}.$ If $n$ is a positive integer, Prove that

$$\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac{2329}{3600}.$$

please don't refer to the famous  $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.

I am looking a method that doesn't use $\text{“}\pi\text{''}.$

Unfortunately, I know and tried only $\text{“}\pi\text{''}$ method.
I like this kind of numerical challenges. Let us exploit creative telescoping. For any $n\geq 2$ we have
$$ \frac{1}{n^2}-\left[\frac{1}{n}-\frac{1}{n+1}\right]=\frac{1}{n^2(n+1)}\tag{1} $$
$$ \frac{1}{n^2(n+1)}-\left[\frac{1}{2n^2}-\frac{1}{2(n+1)^2}\right]=\frac{1}{2n^2(n+1)^2}\tag{2} $$
$$ \frac{1}{2n^2(n+1)^2}-\left[\frac{1}{6n^3}-\frac{1}{6(n+1)^3}\right]=-\frac{1}{6n^3(n+1)^3}\tag{3} $$
$$ \frac{1}{6n^3(n+1)^3}-\left[\frac{1}{30n^5}-\frac{1}{30(n+1)^5}\right]=\frac{5n^2+5n+1}{30n^5(n+1)^5}\tag{4}$$
$$ \frac{5n^2+5n+1}{30n^5(n+1)^5}&lt; \frac{1}{42 n^7}-\frac{1}{42(n+1)^7}\tag{5} $$
hence by setting $g(n)=\frac{1}{n}+\frac{1}{2n^2}+\frac{1}{6n^3}-\frac{1}{30n^5}$ we have
$$\begin{eqnarray*} \zeta(2)-1=\sum_{n\geq 2}\frac{1}{n^2} &amp;=&amp; \sum_{n\geq 2}\left[g(n)-g(n+1)\right]+\sum_{n\geq 2}\frac{5n^2+5n+1}{30n^5(n+1)^5}\\ &amp;&lt;&amp; g(2)+\frac{1}{42\cdot 2^7}=\color{red}{\frac{5779}{8960}}&lt;\frac{2322}{3600}.\tag{6}\end{eqnarray*} $$

The same technique proves Stirling's inequality and the "acceleration formula"
$$ \sum_{n\geq 1}\frac{1}{n^2} = \sum_{n\geq 1}\frac{3}{n^2\binom{2n}{n}}\tag{7} $$
leading to the more accurate approximation $\zeta(2)-1\approx \frac{51077}{79200}$.



There also is a simple and accurate lower bound. By choosing $h(x)=\frac{6x-3}{6x^2-6x+2}$ we have $\frac{1}{x^2}\geq h(x)-h(x+1)$ for any $x\geq 2$, with the difference behaving like $\frac{1}{9x^6}$. It follows that
$$ \zeta(2)-1\geq \color{red}{\frac{9}{14}} = h(2).\tag{8}$$
By selecting $k(x)=\frac{60x^2-60x+31}{60x^3-90x^2+66x-18}$ we get $\zeta(2)-1\leq \frac{151}{234}$, simpler but less accurate than $(6)$.
The simplest approach is to approximate the sum by the integral. The crude way of doing it just uses the fact that $1/x^2$ is a decreasing function, so the integral from $k$ to $k+1$ is bigger than $1/(k+1)^2.$ to make it sharper, one notes that since $1/x^2$ is convex, the integral is actually bigger than $\frac12(1/n^2 + 1/(n+1)^2).$