if $\lim_{n\to\pm\infty} |n|^k\widehat{f}(n)=0$ then $f \in C^\infty$ (in $L^1(\mathbb T)$

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So I wish to prove:

For $f\in L^1(\mathbb T)$:

a) If $f\in \mathcal C^\infty(\mathbb T)$ then $$(1)\displaystyle \lim_{n\to\pm\infty} |n|^k\widehat{f}(n)=0, \forall k\in\mathbb N$$

b) If $(1)$ holds then $f$ coincides a.e. with a function in the class $\mathcal C^\infty(\mathbb T)$.

And I had no issue proving a), there're even many answers to it on here. However b) is the one that is giving me trouble... A Hint was given for b): show first that if $(1)$ holds then we have $\displaystyle \sum_{n\in\mathbb Z} |n|^k|\widehat{f}(n)|<\infty$, for every $k\in\mathbb N$ and then use the Weiestrass theorem and induction, where the theorem says:

Theorem (Weiestrass): Let $\{g_n\}_n$ be a sequence of $\mathcal C^1$ functions defined in an interval of $\mathbb R$ so that $ \sum_{n} \|g_n\|_\infty<\infty$ and $ \sum_{n} \|g_n'\|_\infty<\infty$. Then the series $ \sum_{n} g_n(x)$ converges to a function $g\in \mathcal C^1$ such that $g'(x)= \sum_{n} g_n'(x)$

I have the idea that the series convergence derives from proving the sequence of partial sums is Cauchy but I can't seem to prove it..

Also, is it really called Weierstrass Theorem? No matter how I searched for it I couldn't find this specific result!

Anyways thank you very much!