Easy question about finite energy due to convergence The infinite-length sequence $x_1[n]$ defined by 
\begin{multline}
x_1[n]=
\begin{cases}
\dfrac{1}{n}& \text{if $n \geq $1},\
0& \text{if $n \leq $0}.
\end{cases}
\end{multline}
has an energy equal to 
$\mathcal{E _x {_1}} = \sum^\infty_{n=1}(\dfrac{1}{n})^{2}$
which converges to $\pi^2/6$ indicating that $x_1[n]$ has finite energy.
I don't get where we find $\pi^2/6$. It would be great if anyone can help me out.

Since you are into signal processing, you might like the proof using Parseval's theorem. I have paraphrased this into the signal processing language from the original proof #4 of Robin Chapman's collection of proofs.
Let $e_n = e^{2\pi inx}$ where $n \in \mathbb Z$. Let $f(x) = x$ in the interval $[0,1]$ and we compute its Fourier series
$$ f (x) = \sum_n a_n e^{2\pi inx}.$$
Now, in your terms, Parseval's theorem would mean that the energy computation in the time domain is identical to the energy computation in the frequence domain. To compute energy in the time domain, we integrate the square of the abs. value of the function, and to compute energy in the Frequency domain, we sum the squares of the abs. values of the Fourier coefficients. So,
$$ \int_0^1 x^2 dx\ =\ \sum |a_n|^2$$
As R. Chapman remarks, the left side is $1/3$, and we have $a_0 = 1/2$ and $a_n = 1/2\pi in$ for $n \neq 0$. So the above simplifies to 
$$ \frac{1}{3} = \frac{1}{4} + \underset{n\in \mathbb Z , n \neq 0}{\sum} \frac{1}{4\pi n^2}$$
from which the result follows.

Incidentally, note here that the explicit calculation was not necessary to prove finite energy. The much simpler way is to note that your signal can be bounded by the Fourier series of some other signal, and then observe that energy of that signal as seen in the time domain is finite.
