Proving that $\mathbb E\left[\sum\limits_{k=1}^\infty Z_k\frac{\sqrt 2\sin(k\pi t)}{k \pi}\right] = 0$ (for i.i.d. $Z_k\sim N(0,1)$) I want to prove that that $\mathbb E\left[\sum\limits_{k=1}^\infty \sqrt 2 \frac{Z_k \sin(k\pi t)}{k \pi}\right] = 0$ (for i.i.d. $Z_k\sim N(0,1)$). The way I was thinking to do it was to just interchange integrals and sums and get that
$$\mathbb E\left[\sum\limits_{k=1}^\infty \sqrt 2 \frac{Z_k \sin(k\pi t)}{k \pi}\right] =\sum\limits_{k=1}^\infty \sqrt 2 \frac{\sin(k\pi t)}{k \pi} \mathbb E\left[Z_k\right] = 0$$
However, we need to justify why we can move the integral (i.e. expectation) and sum around. Naturally, one would turn to Fubini-Tonelli, but I do not think that the following is true:
$$\sum_{k=1}^\infty \mathbb E\left[\left|\sqrt 2 \frac{Z_k\sin(k\pi t)}{k \pi}\right|\right] = \sum_{k=1}^\infty \left| \sqrt 2 \frac{\sin(k\pi t)}{k \pi} \right| \mathbb E\left[| Z_k|\right] < \infty$$
Can anyone prove this?
Sidenote: the problem is from proving that $\mathbb U(t,\omega):= \sum\limits_{k=1}^\infty \sqrt 2 \frac{Z_k \sin(k\pi t)}{k \pi}$ is a Brownian bridge. I searched for the Karhunen-Loeve decomposition but it did not provide any help for calculating expectation directly.
 A: Fubini-Tonelli doesn't work here because pulling the absolute value under the infinite sum gives a too rough upper bound. 
The trick is to prove convergence of the partial sums in $L^1$. Set $$S_n := \sum_{k=1}^n \sqrt{2} \frac{\sin(k \pi t)}{k \pi} Z_k.$$ Since the random variables $Z_k$ are independent and have mean zero, we have \begin{align*} \mathbb{E}((S_n-S_m)^2) &= \mathbb{E} \left( \left[ \sum_{k=m+1}^n\sqrt{2} \frac{\sin(k \pi t)}{k \pi} Z_k\right]^2 \right)  \\ &= \sum_{k=m+1}^n \mathbb{E} \left( \left[ \sqrt{2} \frac{\sin(k \pi t)}{k \pi} Z_k \right]^2 \right) \\ &\leq \frac{2}{\pi^2} \sum_{k=m+1}^n \frac{1}{k^2} \end{align*} for all $m<n$. This shows that $(S_n)_{n \in \mathbb{N}}$ is a Cauchy sequence in $L^2(\mathbb{P})$ and, hence also in $L^1(\mathbb{P})$. By completeness of $L^1(\mathbb{P})$, the limit $S:=\sum_{k=1}^{\infty} \sqrt{2} \frac{\sin(k \pi t)}{k \pi} Z_k = \lim_{n \to \infty} S_n$ exists and $\mathbb{E}|S_n-S| \to 0$. In particular, $\mathbb{E}(S_n) \to \mathbb{E}(S)$. From the linearity of the expectation and $\mathbb{E}(Z_k)=0$, it is immediate that $\mathbb{E}(S_n)=0$ for all $n$ and so $\mathbb{E}(S)=0$.
