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Problem:

Prove If the sequence of numbers $\{a_{n}{\}}_{n=1}^{\infty}$ satisfies $\sum_{n=1}^{\infty} \lvert a_n \rvert \lt \infty,$ then the series $\sum_{n=1}^{\infty} a_ncos(nx)$ converges uniformly on $[0,2\pi]$. This means, the partial sums $$ s_N(x) = \sum_{n=1}^N a_ncos(nx) $$ define a sequence of functions $\{s_{N}{\}}_{n=1}^{\infty}$ that converges uniformly on $[0,2\pi]$. Hint: First show that the sequence is Cauchy w.r.t $\lVert \cdot \rVert_{\infty}$

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My thoughts:

I think I need to use the following property -

Property: If a sequence $\{f_{n}{\}}_{n=1}^{\infty}$ in $C([a,b])$ is Cauchy w.r.t $\lVert \cdot \rVert_{\infty}$ (uniformly cauchy) then it is uniformly convergent.

To be honest, the above (and hint) is the part I'm stuck on.

For just $g(x) = \sum_{n=1}^{\infty} cos(nx)$ I feel like I could do this sort of proof.

W.L.O.G., let $k \lt L$

Now: $\lvert \sum_{n=1}^{L} g_n(x) - \sum_{n=1}^{k} g_n(x) \rvert = \lvert \sum_{n=k+1}^{L} g_n(x) \rvert \le \int_k^L 1 \ dx \lt x \ \Big|_k^L$ = L - k

Anyway, I'm probably way off course at this point and I'm having a hard time actually figuring out how to go about this.

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You should consider $|s_{n}(x)-s_{n+p}(x)|\leq\displaystyle\sum_{k=n+1}^{n+p}|a_{k}|$.

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This is just the Weierstrass $M$-test which states that $\sum f_n(x)$ converges uniformly on a set $E$ if $|f_n(x)|\le M_n$ for all $x\in E$ and $\sum M_n < \infty$. In this case, $|a_n\cos(nx)|\le |a_n|$ and $\sum |a_n| < \infty$.

To prove the $M$-test, you can prove that $\{s_N = \sum_{n=1}^Nf_n(x)\}_{N=1}^\infty$ is Cauchy by looking at $|\sum_{n=m}^p f_n(x)| \le \sum_{n=m}^p|f_n(x)| \le \sum_{n=m}^p M_n$, and use the fact that $\sum M_n < \infty$, so in particular $\{\sum_{n=1}^NM_n\}_{N=1}^\infty$ is Cauchy.

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