How can I interchange the sum signs I need to interchange the sum signs in the following double summation
$$\sum_{n=1}^{\infty}\sum_{k\in\mathbb{Z}^{*}}\hat{g}(k)e^{2\pi ink\theta}n^{-s},$$
where $\mathbb{Z}^{\star}$ is the set of integers numbers excluded 0, $\hat{g}(k)$ is the Fourier coefficient of an analytic, 1 periodic function, real function, (so $|\hat{g}(k)|\leq Ck^{-\delta}$, for every $\delta>0$), $\theta$ is an irrational number and $s$ a complex number with $\sigma=\mathrm{Re}(s)>0$. The problem is, I can't prove that the double series
$$\sum_{n=1}^{\infty}\sum_{k\in\mathbb{Z}^{*}}|\hat{g}(k)e^{2\pi ink\theta}n^{-s}|=\sum_{n=1}^{\infty}\sum_{k\in\mathbb{Z}^{*}}|\hat{g}(k)|n^{-\sigma}$$
converge. It seems, that since $|\hat{g}(k)|\leq Ck^{-\delta}$, for every $\delta>0$, I need to find a sequence $A(n)$ such that $|\hat{g}(k)|<k^{-A(n)}$, this is justified by the condition on $\hat{g}$, and $$\sum_{k\in\mathbb{Z}^*}k^{-A(n)}<Cn^{-1},$$
where $C$ is a positive constant. Then $$\sum_{n=1}^{\infty}\sum_{k\in\mathbb{Z}^{*}}|\hat{g}(k)|n^{-\sigma}<C\sum_{n=1}^{\infty}n^{-1-\sigma}<\infty,$$
by the integral criterion. Does a sequence like $A(n)$ exist? Provided that sequence, is my argument correct? How can I find a sequence like this.
 A: Apart from symbols, and supposing that I understood properly what you mean to achieve, let me present this just as a hint (too long for a comment).
If the $g(k)$'s are the Fourier coefficients, then they are tied to the amplitude (let's call them $A(k)$) of the harmonics composing the signal. The relating coefficient depends on how the Fourier series is defined.
Since you say that the originating signal is a real periodical function, then its power is limited (and maybe you can calculate it).
If it is so, then you know that the sum of the squares of the $A(k)$ is limited, and that should help you in carrying on the demonstration.
That is, if
$$
f(x) = f(x + 1) = {{A_{\,0} } \over 2} + \sum\limits_{1\, \le \,k} {A_{\,k} \cos \left( {2k\pi \,x + \varphi _{\,k} } \right)}  = \sum\limits_{ - \;\infty \, < \,k\, < \,\infty \;} {c_{\,k} \;e^{\,i2k\pi \,x} } 
$$
with
$$
\left\{ \matrix{
  {{A_{\,0} } \over 2} = c_{\,0}  \hfill \cr 
  \left| {c_{\,k} } \right|\; = \left| {c_{\, - \,k} } \right| = {{A_{\,k} } \over 2} \hfill \cr 
  \varphi _{\,k}  = \arg \left( {c_{\,k} } \right)\quad \left| {\;1 \le k} \right. \hfill \cr}  \right.
$$
then
$$
\int_{x = 0}^{\,1} {f(x)^{\,2} dx}  = {{A_{\,0} ^{\,2} } \over 4} + {1 \over 2}\sum\limits_{1\, \le \,k} {A_{\,k} ^{\,2} } 
$$
Which is the Power Theorem or Parseval's theorem
