Prove that for every $n \in \mathbb{N}$ $\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$ $$\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$$


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*First step would be proving that the statement is true for n=2


On the LHS for $n=2$ we would have $\frac{1}{4}$  therefore the statement is true for $n=2$


*Now we must assume the statement is true for $n=j$ with $j\geq2$
$$\sum\limits_{k=2}^{j}{\frac{1}{k^2}}<1$$


*Now we must prove true for $n=j+1$
$\sum\limits_{k=2}^{j+1}{\frac{1}{k^2}}<1$
$\Rightarrow \sum\limits_{k=2}^{j}{\frac{1}{k^2}}+\frac{1}{(k+1)^2}<1$
I cant seem to introduce the induction hypothesis 

$f(x)= \frac{1}{x^2}$, $x >0$, stricly decreasing.
$ \displaystyle{\sum_{k=2}^{n}}\frac{1}{k^2} <\displaystyle{ \int_{1}^{n}}\frac{1}{x^2}dx =-\frac{1}{x}\big ]_1^n=$
$1-1/n. $
As suggested by Calvin you can use the above inequality for the induction proof.
