Why is it that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ is not less than $\int_1^\infty \frac{dx}{x^2} = 1$? So according to Euler's proof of the Basel problem, 
$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6},$$
But only for $n  \in  \mathbb{Z}$.
But if $n$ was a positive real and $n \geqslant 1$, then would the sum $S$ be equal to,
$$\int_1^\infty \frac{dx}{x^2}?$$
If so then after solving the integral via power rule we get $S=1$.
But how can this be since when taking reals, we take integral values as well as new values. So by logic shouldn't, $S>\dfrac{\pi^2}{6}$ but $1$ isn't. Where am I going wrong pls explain.

Why should the integral be equal to the sum? To see the sum must be larger,
note that
$$\int_1^\infty\frac{dx}{x^2}=\sum_{n=1}^\infty
\int_n^{n+1}\frac{dx}{x^2}<\sum_{n=1}^\infty\frac1{n^2}$$
as $x\mapsto 1/x^2$ is a decreasing function.
