<s>[INST] Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $  \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $
my problem is can we prove the statement 
"for each $n\in \Bbb N$ , $  \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $"
only using induction ?(without using the fact that convergence of the above series)
any ideas, thanks!
 [/INST] Take a look at this proof of Daniel J. Velleman. 
For $n\geq 1$, we define the positive numbers,
$$I_n:=\int_0^{\pi/2}(\cos(x))^{2n}\,dx\quad\text{and}\quad
J_n:=\int_0^{\pi/2}x^2(\cos(x))^{2n}\,dx.$$
By integration by parts we show that
$$I_n=(2n-1)(I_{n-1}-I_n)\implies I_n=\frac{2n-1}{2n}I_{n-1}$$
and 
$$I_n=n(2n-1)J_{n-1}-2n^2J_n.$$
Hence, by dividing the last one by $n^2I_n$, we get
$$\frac{1}{n^2}=\frac{(2n-1)J_{n-1}}{nI_n}-\frac{2J_n}{I_n}=
2\left(\frac{J_{n-1}}{I_{n-1}}-\frac{J_n}{I_n}\right).$$
It follows by induction (see the telescopic sum) that for $K\geq 1$,
$$\frac{\pi^2}{6}-\sum_{n=1}^K\frac{1}{n^2}=\frac{\pi^2}{6}-2\sum_{n=1}^K\left(\frac{J_{n-1}}{I_{n-1}}-\frac{J_n}{I_n}\right)=\frac{\pi^2}{6}-2
\left(\frac{J_0}{I_0}-\frac{J_K}{I_K}\right)=\frac{2J_K}{I_K}>0$$
where we used that $I_0=\pi/2$ and $J_0=\pi^3/24$.
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