Calculating the norm of an infinite vector I'm reading "Introduction to Hilbert Spaces" by N. Young. Right in the first chapter, after introducing inner products and norms in general linear spaces, it asks to show that the norm of the vector:
$$x = \left(\frac{1}{n}\right)_{n=1}^\infty$$
is $\frac{\pi}{\sqrt 6}$, so basically:
$$\langle x,x \rangle= \sum_1^\infty \frac{1}{n^2}$$
must be $\frac{\pi^2}{6}$
I know this is generally done by using the Fourier transform of $x^2$ to calculate the value of $\zeta(2)$, which is that. This however, appears just as the beginning of the book, while Fourier analysis is supposed to be a part of the course that comes later, and so I imagine there must be a simpler more direct way of calculating that, specially because it's among the exercises you solve in less than a line. I've found other ways of calculating the zeta function in a page of wolfram, but is there any special way for this particular value that is partuclarly easy and that I don't know?
Thank you.
 A: Note that,
$$
1 - \frac{\pi^2}{6} z^2 + \ldots = \frac{\sin(\pi z)}{\pi z} = \prod_{n > 0} \bigg ( 1 - \frac{z^2}{n^2} \bigg ) = 1
- \bigg ( \sum_{n > 0} \frac{1}{n^2} \bigg)z^2 + \ldots 
$$
Comparing the coefficients of $z^2$ we get the claim.
Details: The first equality is just the Taylor serie for $\sin(\pi z)/(\pi z)$ obtained by differentiating. The second equality follows from complex analysis, and is established by noticing that the two functions vanish at exactly the same points and are both of not "high growth", the last equality is established combinatorially by expanding the product and looking at the coefficient of $z^2$. 
More details on the second equality: Let $f(z) = \sin (\pi z)$ and $$g(z) = \pi z \prod_{n > 0} \bigg ( 1 - \frac{z^2}{n^2} \bigg ).$$
Notice that $|f(z + 1)| = |f(z)|$ and $|g(z + 1)| = |g(z)|$ (*). 
Let
$$
H(z) = \frac{g(z)}{f(z)}
$$
Notice that $|H(z + 1)| = |H(z)|$. Furthermore $H(z)$ is entire, and has no zeros, therefore $H(z) = e^{P(z)}$ with $P$ analytic. One can easily show that $g(z) = O(e^{c |z|^2})$ for large $|z|$, therefore $P$ must be a polynomial of degree at most $2$. But since $|H(z + 1)| = |H(z)|$ we have $\Re P(z + 1) = \Re P(z)$ for all $z$, hence $P$ is a constant. Since $H(0) = 1$ it follows that $P = 0$. Hence $H(z) = 1$, and so $f(z) = g(z)$ as desired. 
(*): to prove this note that $\frac{g'(z)}{g(z)} = \sum_{n \in \mathbb{Z}} \frac{1}{n - z}$ so $\frac{g'(z + 1)}{g(z + 1)} = \frac{g'(z)}{g(z)}$ integrating and taking the real part we get $\log |g(z + 1)| = \log |g(z)|$ (this was actually somewhat tricky to do, I took a rectangular contour around $z =n$, with one side of length $1$, the contribution of two of the sides cancel, then I took the real part and used the fact that $|g(x + iy)| = |g(x - iy)|$... It was more annoying than I thought so maybe you have a better method...
