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Is the series \( \displaystyle \sum_{n=1}^{\infty} \frac{\cos n x}{\ln (n+2)} \) a Fourier series of some \( 2\pi \) periodical function?

The answer would be no due to Parsheval's identity. Indeed:

$$\sum_{n=1}^{\infty}a_n^2 =\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)\,{\rm d}x \Leftrightarrow \sum_{n=1}^{\infty}\frac{1}{\ln^2 (n+1)}=\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)\,{\rm d}x$$

which is a contradiction, since the series $\sum \limits_{n=1}^{\infty} \frac{1}{\ln^2 (n+1)}$ is known to diverge.

The sequence is convex.The series is Fourier series of Lebesgue integrable function.

S.F.Papadopoulos wrote:The sequence is convex.

I get the thing with the Fourier series and Lebesque but I do not get what do you mean by "the sequence is convex".

If $2a_{n+1}<a_n+a_{n+2}$ for all natural $n$ the sequence is strictly convex.