Consider
Suppose that an analytic function $f$ on $[-1,1]$ is not periodic, yet $f(-1)=f(+1)$ and $f^{\prime}(-1)=f^{\prime}(1)$. Integrating by parts the Fourier coefficients $\hat{f}_n$ show that $\hat{f}_n=\mathcal{O}\left(n^{-3}\right)$. Show that the rate of convergence of the $N$-terms truncated Fourier expansion of $f$ is hence $\mathcal{O}\left(N^{-2}\right)$.
Simple integration by parts and using the assumptions on the end points yielded me the first part. However, I am strugling to see how to argue for the second part. In particular, I know that I need to show that $$ \sum_{|n|>N/2} \hat f_n e^{i \pi n x}=\mathcal{O}(N^{-2}) $$ but my issue is that the sum here is infinite. I can kind of see where this is coming from, "as per dodgy analysis", $$ \sum_{|n|>N/2} \hat f_n e^{i \pi n x}= \sum \mathcal{O}(N^{-3}) =N \cdot\mathcal{O}(N^{-3})= \mathcal{O}(N^{-2}). $$ Obviously, this, as written is nonsense, but I am struggling to even come close to getting a sound argument.
Moreover, I am hesitant to use $| \,|$ and then the triangle inequality as then I will get an inequality of the sort $$ |\sum_{|n|>N/2} \hat f_n e^{i \pi n x} |\leq g =\mathcal{O}(N^{-2}) $$ which certainly does not imply the result.
For context, this question comes from a course on numerical analysis, so perhaps I can impose some further restrictions on $f$ and just assume that the question was ill stated?
Question: How do I deal with infinite Fourier sums in the context of strict bounding as is the question?
The Fourier series is absolutely convergent as $|\hat{f}_n|=\mathcal{O}(n^{-3}).$ Moreover we have $$ \left |\sum_{|n|>N} \hat f_n e^{i \pi n x}\right |\le \sum_{|n|>N} |\hat f_n |\le C\sum_{|n|>N} |n|^{-3}=\mathcal{O}(N^{-2}) $$