Let $f_n(x)=1+\sin(n\pi x)$. I have to prove that $f_n^k$ converges weakly in $L^1([0,1])$ to \begin{equation} c_k:=\underset{a+2b=k}{\sum_{a,b\geq0}}\frac{k!}{a!b!b!}4^{-b} \end{equation} for $n\rightarrow+\infty$.
I cannot use Riemann Lebesgue lemma, so I don't know how to do this. Could somebody give me a hint?
Hint. Write
\begin{align*} f_n^k &= \left( 1 + \frac{i}{2i}e^{n\pi ix} - \frac{1}{2i}e^{-n\pi ix}\right)^k \\ &= \sum_{\substack{a,b,c\geq 0 \\a+b+c=k}} \binom{k}{a,b,c} \left(\frac{1}{2i}e^{n\pi ix}\right)^b\left( -\frac{1}{2i}e^{-n\pi ix}\right)^c \\ &= \sum_{\substack{a,b,c\geq 0 \\a+b+c=k}} \binom{k}{a,b,c} \frac{1}{2^{b+c}i^{b-c}} e^{n\pi(b-c)ix} \end{align*}