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How to show that $$ \sum_{n=2}^\infty \frac{\sin{(nx)}}{\log n} $$ not the Fourier series of any function?

I have shown that the series is convergent by Dirichlet test.

Let $a(n)=\frac{1}{\log n}$. What is $\sum (a(n))^2$, to apply Parseval's theorem?

t.b.
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jr92
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2 Answers2

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Since $\Bigl\lvert\sum\sin(n x)\Bigr\rvert\leq M$ for $M<\infty$ and $\Bigl\{\frac{1}{\log(n)}\Bigr\}\to 0$ monotonically, by Dirichlet's test, $\sum_{n=2}^{\infty}\frac{\sin(nx)}{\log(n)}$ converges.

If $\sum_{n=2}^{\infty}\frac{\sin(nx)}{\log(n)}$ is a Fourier series, by Parseval's theorem, there exist a Riemann integrable function $f$ such that $$ \int_{-\pi}^{\pi}\lvert f(x)\rvert^2dx=2\pi\sum_{n=2}^{\infty}\frac{1}{\log^2\lvert n\rvert} $$

Theorem:

Suppose $a_1\geq a_2\geq\cdots\geq 0$. Then the series $\sum a_n$ converges iff $\sum 2^ka_{2^k}$ converges. $$ \sum_{n=2}^{\infty}\frac{2^k}{k^2\log^2(2)}\geq\sum_{n=2}^{\infty}\frac{1}{k}=\infty $$

Therefore, by Cauchy's condensation test, the $\sum_{n=2}^{\infty}\frac{1}{\log^2\lvert n\rvert}$ does not converge which contradicts Parseval's theorem. Thus, $\sum_{n=2}^{\infty}\frac{\sin(nx)}{\log(n)}$ is not a Fourier series.

dustin
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Suppose that it is the Fourier series of $f(x)$ in $(-\pi,\pi)$, namely, $$ f(x)=\sum_{n=2}^\infty \frac{\sin(nx)}{\log n},\quad x\in(-\pi,\pi).$$ Then by Pareraval's Identity, $$\int_{-\pi}^\pi f(x) \, dx=\sum_{n=2}^\infty\frac{1}{\log^2n}=\infty.$$

Michael Hardy
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xpaul
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    This answer is incomplete. Parseval's identity applies when $\int_{-\pi}^\pi |f(x)|^2,dx<\infty$. What about the case when that integral is infinite but $\int_{-\pi}^\pi |f(x)| , dx<\infty$? Those functions also have Fourier series. – Michael Hardy Mar 11 '13 at 18:31
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    . . . also, did you intend $\int_{-\pi}^\pi |f(x)|^2,dx$ where you wrote $\int_{-\pi}^\pi f(x),dx$? – Michael Hardy Mar 11 '13 at 18:33
  • The proof is correct if we assume f is Riemann integrable though... – XIIIX Mar 21 '20 at 09:58