<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] The statement that we will prove with induction is that for every $K > k^*$, 
$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 - \frac 1K $$
where $k^* = 4091641$. 
This implies that, for every $K$,
$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 $$

The idea is to find a function $f(K) > 0$ such that 
$$\sum_{n=1}^K \frac {1}{n^2} \le \frac {\pi ^2} 6 - f(K) $$
As @orlp states in his comment, it is clear that $f(K)$ cannot be a constant, because we know that the partial sums get arbitrarily close to $\frac{\pi^2}6$. So which function do we pick?
Well, just write the inductive step:
$$\sum_{n=1}^K \frac {1}{n^2} + \frac{1}{(K+1)^2} \le \frac {\pi ^2} 6 - f(K) + \frac{1}{(K+1)^2} \mathop{\le}^?  \frac {\pi ^2} 6 - f(K+1)$$
If we can prove the last inequality, then the inductive step will work and we will have solved the problem.
The last inequality is equivalent to 
$$f(K+1) \le f(K) - \frac 1{(K+1)^2}$$
which means that $f(K)$ must be decreasing. One can easily check that $f(K) = \frac 1K$, for example, works. The only thing that is left is the base case. 
Turns out, the base case is not so easy to find. A quick numerical simulation, though, proved that for $n=4091641$ the case case is satisfied, and the rest follows. Note that since we know that the statement is true for a certain $K > k^*$, we also know that every partial sum up to $k^*$ is $\le \frac{\pi^2}6$, as the partial sums are increasing.
One could probably find a better $f(K)$ such that we don't need computers to verify the base case, but I'll leave that to someone else :)
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