$L^2$ norm of series with complex exponentials Suppose that $f\in L^2(\mathbb{R}/2\pi\mathbb{Z})$ takes the form $$f(\theta)=\sum_{n=1}^\infty a_ne^{in\theta}.$$ The function $$f_r(\theta)=\sum_{n=1}^\infty r^na_ne^{in\theta}$$ is a harmonic function in $|z|<1$. Then 
$$\|f_r-f\|_2=\int_\mathbb{R}\left|\sum_{n=1}^\infty(1-r^n)a_ne^{in\theta}\right|^2d\theta.$$
I want to show that $\lim_{r\rightarrow 1^-}\|f_r-f\|_2=0$. But the formula above for $\|f_r-f\|_2$ is a mess. Can I simplify it somehow?
 A: Not such a big mess, since the functions $e^{in\theta}/(2\pi)$ form an orthonormal system in $L^2(\mathbb R/2\pi \mathbb Z)$. 
$$ \frac{1}{(2\pi)^2}\int_{-\pi}^{\pi}  \left|\sum_{n=1}^\infty(1-r^n)a_ne^{in\theta}\right|^2\,d\theta =  \sum_{n=1}^\infty (1-r^n)^2 |a_n|^2$$ 
Termwise, the series converges to $0$. It is also dominated by $\sum_{n=1}^\infty   |a_n|^2$, which is finite. By the Dominated Convergence theorem (applied to $\mathbb N$ with the counting measure), 
$$\sum_{n=1}^\infty (1-r^n)^2 |a_n|^2\to 0$$
Details added upon request: 


*

*Consider the measure space $X=(\mathbb N, 2^{\mathbb N}, \mu)$   where $\mu(A)$ is the number of elements of $A$.

*The sequence $(|a_n|^2)$ is in $L^1(X)$. 

*For every $r\in [0,1)$ we have $(1-r^n)^2|a_n|^2 \le |a_n|^2 $ for all $n$; this is domination. 

*For every $n$, we have pointwise convergence $(1-r^n)^2|a_n|^2 \to |a_n|^2 $ as $r\to 1$. 

*Apply the dominated convergence theorem based on 2-3-4. Since it's usually stated for sequences of functions, let $r$ run  through a sequence $(r_k)$ converging to $0$. Since the sequence $r_k$ is arbitrary, and the limit is the same, the conclusion follows. 


Alternative version, with proof from basic principles (no  DCT). 
Given $\epsilon>0$, pick $N$ such that $\sum_{n>N}|a_n|^2<\epsilon/2$. 
Since for every $n$ we have $(1-r^n)^2|a_n|^2\to 0$ as $r\to 1$, it follows that 
$$\sum_{n=1}^N (1-r^n)^2|a_n|^2 \to 0\quad \text{as } \ r\to 1 \tag{1}$$
Thus, there exists $\delta>0$ such that the left side of (1) is less than
$\epsilon/2$ whenever $r>1-\delta$. 
Conclusion: 
$$\sum_{n=1}^\infty (1-r^n)^2|a_n|^2 <\epsilon \quad\text{whenever } \ 
 r> 1-\delta  $$
