Showing that inner product of two vectors is the limit of the inner products How can you show that 
$$(a,b) = (\sum_{i=1}^\infty a_i \phi_i,\sum_{i=1}^\infty b_i \phi_i) = \lim_{N\to \infty}(\sum_{i=1}^N a_i \phi_i,\sum_{i=1}^N b_i \phi_i)$$
where $\phi_i$ is an orthonormal basis in a Hilbert space and $a$ and $b$ are two vectors in it with components $a_i$ and $b_i$ respectively?
 A: Going to an orthonormal basis actually obfuscates things a bit. There's no need for it and it can be misleading. Let $a,b\in\mathcal{H}$, $a_n\rightarrow a$, and $b_n\rightarrow b$. That is to say that given $\varepsilon > 0$, there exists $N$ such that for all $n>N$, $||a_n-a||<\varepsilon$ and $||b_n-b||<\varepsilon$.
Let's observe what happens to $\langle a_n,b_n\rangle$. Let $\varepsilon > 0$ and $n>N$ such that $||a_n-a||<\varepsilon$ and $||b_n-b||<\varepsilon$.
$$|\langle a_n,b_n\rangle - \langle a,b\rangle| = |\langle a_n,b_n\rangle - \langle a_n,b\rangle + \langle a_n,b\rangle -\langle a,b\rangle|$$
We then have (using the triangle inequality for real numbers):
$$|\langle a_n,b_n\rangle - \langle a,b\rangle| \le |\langle a_n,b_n\rangle - \langle a_n,b\rangle| + |\langle a_n,b\rangle -\langle a,b\rangle|.$$
Making use of the linearity of the inner product:
$$|\langle a_n,b_n\rangle - \langle a,b\rangle| \le |\langle a_n,b_n-b\rangle| + |\langle a_n-a,b\rangle|.$$
If we use Cauchy-Schwarz we get
$$|\langle a_n,b_n\rangle - \langle a,b\rangle| \le ||a_n||\,||b_n-b|| + ||a_n-a||\,||b||.$$
Now we need to make use of the fact that if a sequence converges it is bounded so that $||a_n||<M$ for some $M$ large enough. Then we have
$$|\langle a_n,b_n\rangle - \langle a,b\rangle| < M\varepsilon + ||b||\varepsilon.$$
You can easily rescale the convergence criteria for $a_n,b_n$ so that the right hand side is less than $\varepsilon$ but this is sufficient.
To tie this back to your problem, let $a_n,b_n$ be the partial sums in your orthonormal basis. Provided you have shown these converge, the above analysis gives you exactly what you want.
A: First prove that if $x_n \rightarrow x$ and $y_n \rightarrow y$ then $\langle x_n,y_n\rangle\rightarrow \langle x,y\rangle$, where the convergence considered for the first two arrows is that one induced by the inner product via the norm $\|\;\| = \langle\,,\rangle$. This says that the inner product is a continuous function of its arguments.
Take $\{\phi_i\}_{i\in\mathbb{N}}$ an orthonormal basis and put, given $a$ in your Hilbert space, $a_i=\langle a,\phi_i \rangle$ for every $i\in \mathbb{N}$. Then you have:


*

*$\sum_{i=1}^\infty a_i \phi_i$ converges because
$\{\phi_i\}_{i\in\mathbb{N}}$ is an orthonormal set

*$\sum_{i=1}^\infty a_i \phi_i$ converges to $a$ because
$\{\phi_i\}_{i\in\mathbb{N}}$ is also complete


With these two things you get the first equality of your question because $\sum_{i=1}^\infty a_i \phi_i$ and $\sum_{i=1}^\infty b_i \phi_i$ are just rewrites of $a$ and $b$ respectively. Using the first observation of this answer, and the definiton of what a series is, you get the last equality.
A: The inner product is continuous, this follows immediately from the polarization identity (http://en.wikipedia.org/wiki/Polarization_identity).
Hence if $a_n \to a$, and $b_n \to b$, then $\langle a_n, b_n \rangle \to \langle a, b \rangle $.
Let $a_n = \sum_{i=1}^n a_i \phi_i$ and $a = \sum_{i=1}^\infty a_i \phi_i$, and similarly for $b_n,b$ to get the desired result.
