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.
 A: A viable approach is also to apply a "delayed" creative telescoping based on the identity $\frac{1}{n^2}<\frac{1}{n^2-1}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)$. This approach leads to
$$ \zeta(2)-1<\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}\right)+\frac{1}{2}\sum_{n>5}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)=\frac{2329}{3600} $$
in a single line.
A: Let $\displaystyle S = \sum_{n=2}^\infty \frac 1 {n^2}.\,$ A simple integral test shows $S<\infty.$
Let $T$ be some number a little bit bigger than $S.$ Then
\begin{align}
& \frac 1 {2^2} + \cdots + \frac 1 {n^2} \\[10pt]
< {} & \frac 1 {2^2} + \cdots + \frac 1 {n^2} + \int_n^\infty \frac{dx}{x^2} \\[10pt]
= {} & \left(\frac 1 {2^2} + \cdots + \frac 1 {n^2} \right) + \frac 1 n \tag 1 \\[10pt]
< {} & S + \frac 1 n \\[10pt]
< {} & T \text{ if $n$ is big enough.}
\end{align}
This shows $\displaystyle \frac 1 {2^2} + \cdots + \frac 1 {n^2} < T$ if $n$ is big enough, but that sum is smaller if $n$ is smaller; therefore that inequality holds without any qualification on the size of $n.$ 
Thus we only need to know how big $T$ needs to be, or how small $T$ can be. I think maybe if you take $n=500$ then line $(1)$ might show that $T=2329/3600$ can serve.
This is computation-intensive. Maybe there's also an intelligent way to do it.
A: For any natural $n$, we have
$$S_1=\sum_{k=2}^\infty\frac1{k^2}>\sum_{k=2}^n\frac1{k^2}$$
Likewise, consider the following alternating sum:
$$S_2=\sum_{k=2}^\infty\frac{(-1)^k}{k^2}$$
And see that we have
$$S_1+S_2=\sum_{k=2}^\infty\frac{1+(-1)^k}{k^2}=\sum_{k=1}^\infty\frac2{(2k)^2}=\frac12+\frac12S_1$$
Thus, we have
$$S_1=1-2S_2$$
Furthermore, it is easy to see that
$$S_2>\sum_{k=2}^{23}\frac{(-1)^k}{k^2}$$
And thus,
$$S_1<1-2\sum_{k=2}^{23}\frac{(-1)^k}{k^2}<\frac{2329}{3600}$$
Not a terrible sum to evaluate there.
A: approximate by telescoping series. 
$\left(\sum_{k=2}^{20} \dfrac1{k^2}\right)+(\frac1{20}-\frac1{21})+(\frac1{21}-\frac1{22})+・・・(\frac1{n-1}-\frac1n)=.62976+\frac1{20}-\frac1n=.64616-\frac1n<0.6469\dot4=\frac{2329}{3600}$.
Also $16$ instead of 20 holds.
A: 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 $\frac12(1/n^2 + 1/(n+1)^2).$
