Inner product of an infinite sum For complex numbers $a_{n}, b_{n}$, what would be the next step to simplify this expression:
$$\left \|\sum_{n=1}^\infty a_{n} b_{n}\right \|^{2}=\left \langle \sum_{n=1}^\infty a_{n}b_{n},\sum_{m=1}^\infty a_{m}b_{m}\right \rangle$$
Is it $=\sum_{n=1}^\infty \sum_{m=1}^\infty |a_{n}|^{2}|b_{m}|^{2}$ !? 
where $|a_{n}|^{2}=a_{n}\overline{a_{n}}$
 A: You could simplify it some this way: 
$$\left \langle \sum_{n=1}^\infty a_{n}b_{n},\sum_{m=1}^\infty a_{m}b_{m}\right \rangle=\lim_{N \to \infty} \lim_{M\to \infty} \left \langle \sum_{n=1}^N a_{n}b_{n},\sum_{m=1}^M a_{m}b_{m}\right \rangle = \lim_{N \to \infty} \lim_{M\to \infty} \sum_{n=1}^N \sum_{m=1}^M \left \langle a_{n}b_{n}, a_{m}b_{m}\right \rangle$$ $$ = \sum_{n=1}^\infty \sum_{m=1}^\infty \left \langle a_{n}b_{n}, a_{m}b_{m}\right \rangle=\sum_{n=1}^\infty \sum_{m=1}^\infty a_{n}b_{n} \overline{a_{m}b_{m}}$$
There are some details in between that I didn't want to type, and I'm a little iffy about the convergence, but I think it works.  You can pass the limit outside the inner product because inner products are continuous.  Your expression is missing the cross terms.
A: I don't think you can simplify it any further.  Your simplification does not work because $\sum_{n=1}^\infty \sum_{m=1}^\infty |a_{n}|^{2}|b_{m}|^{2}$ is actually an upper bound for your expression.  To see this, test the example $1*1 + 1*(-1) + \cdots = 0$.
