Let $H\in (0,1/2)\cup (1/2,1)$. For $\lambda\in [-\pi,\pi]$ let $$f_H(\lambda)=\vert e^{i\lambda}-1\vert^2 \sum_{k\in\mathbb{Z}}\vert \lambda+2k\pi\vert^{-2H-1}.$$
I am reading a book that say "it easy to see that $$f_H(\lambda)\sim \vert \lambda\vert ^2\vert \lambda\vert ^{-2H-1}=\vert \lambda\vert^{-2H+1}$$ as $\lambda\to 0.$"
It will follow for exemple from $$\sum_{k\in\mathbb{Z}}\vert 1+\frac{2k\pi}{\lambda}\vert^{-2H-1}\to 1$$ as $\lambda\to 0.$ But I don't see how to prove this kind of limit.
Note that for $\lambda\in [-\pi,\pi]$ and $\alpha>1$, $$ \sum_{k\in\mathbb{Z}}|\lambda+2\pi k|^{-\alpha}=2\sum_{k=1}^\infty(\lambda+2\pi k)^{-\alpha}+|\lambda|^{-\alpha}. $$ By monotone convergence, it is clear that you can take the limit $\lambda\to 0$ in the first sum. Hence, $$ |\lambda|^{-2}|\lambda|^{\alpha} |e^{i\lambda}-1|^2\sum_{k\in\mathbb{Z}}|\lambda+2\pi k|^{-\alpha}=\frac{|e^{i\lambda}-1|^2}{|\lambda|^2}\left(2|\lambda|^{\alpha}\sum_{k=1}^\infty(\lambda+2\pi k)^{-\alpha}+1\right)\to 1 $$ as $\lambda\to 0$.