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Let's consider function $f(x)$ that is $2\pi$-periodic, analytical and $f(x) \ge m > 0$. And consider its Fourier series $f(x) = \sum\limits_{k=-\infty}^{\infty}a_k e^{ikx}$. Then consider $\frac{1}{f(x)}$ and its Fourier series $1/f(x) = \sum\limits_{k=-\infty}^{\infty}b_k e^{ikx}$. Then I claim that $\sum\limits_{k=-n}^{n}b_{-k}a_k$ tends to $1$. How to prove it?

Really I have no idea how to prove it. Maybe apply something like summation by parts? I tested it numerically for some functions - it works. But I really don't know even how to start the proof.

Great thanks for any help or ideas!

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    Have you seen Parseval's identity? – 2017-02-28
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    @Bowditch, how might Parseval be applicable here? – 2017-02-28
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    Multiply together and equate constant terms? – 2017-02-28

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Both $f(x)$ and $g(x)=\frac{1}{f(x)}$ are functions in $L^2(0,2\pi)$. In particular, their (symmetrically) truncated Fourier series are converging to $f(x)$ and $g(x)$ in $L^2$. The term $$ \sum_{n=-N}^{N}a_n b_{-n} $$ is the mean value, over $(0,2\pi)$, of the product between two truncated Fourier series. Since the product of such truncations converges to $f(x)\cdot g(x)=1$ in $L^2$, $$ \lim_{N\to +\infty}\sum_{n=-N}^{N}a_n b_{-n} = \frac{1}{2\pi}\int_{0}^{2\pi}f(x)\,g(x)\,dx = 1 $$ as wanted. It is interesting to apply such lemma to $f(x)=\frac{1}{3+2\cos(x)}$, for instance. In such a case we find that the Fourier coefficients of $f(x)$ are deeply related with the continued fraction of $\sqrt{5}$.

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    Why (symmetrical) truncation here? – 2017-03-01
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    Why restricting to the open interval? – 2017-03-01
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    @FreeziiS: because otherwise $\sum_{n=-N}^{N}a_n b_{-n}$ is not what I told. The choice of the open interval $(0,2\pi)$ is just conventional, here $\int_{(0,2\pi)}=\int_{[0,2\pi]}$. – 2017-03-01
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    Are you saying, e.g.: $\lim_N\sum_{n=-N}^N(\ldots)\neq\lim_N\sum_{n=-N}^{2N}(\ldots)$ – 2017-03-01
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    @FreeziiS: no, I am not saying that. The limit is the same in the $L^2$ sense, but I need that $\sum_{n=-N}^{N}a_n b_{-n}=\frac{1}{2\pi}\int_{0}^{\pi}\tilde{f}(x)\,\tilde{g}(x)\,dx $ where $\tilde{f}$ and $\tilde{g}$ are given by truncations of the Fourier series of $f$ and $g$, so such truncations have to be symmetric. – 2017-03-01