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.