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Let $\{a(n)\}_{n\in\mathbb N}$ be a sequence of real number, suh that for any $C\in \mathbb{R}$ we have $$a(n)\ll_{C}n^{C}$$ My question : is how we can prove that the Dirichlet series $$\sum_{n=1}^{\infty}\frac{a(n)^2}{n^s}$$ converges for all $s\in \mathbb{C}.$

Many thanks

Med
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1 Answers1

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Given an $s\in\mathbb{C}$, we have a $b$ so that $\left|a(n)\right|\le bn^{\mathrm{Re}(s/2)-1}$ (let $C=\mathrm{Re}(s/2)-1$). Then $$ \left|\frac{a(n)^2}{n^s}\right|\le\frac{b^2}{n^2} $$ Therefore, $$ \sum_{n=1}^\infty\frac{a(n)^2}{n^s} $$ converges absolutely by comparison.

robjohn
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  • Mister Johnson, I have a similar question : the sequence ${a(n)}{n\in\mathbb N}$ satisfy $a(n)\ll{\epsilon}n^{k-1+2\epsilon}$ for any $\epsilon>0$ and I wanna prove that the Dirichlet series $\sum_{n=1}^{\infty}\frac{a(n)^2}{n^s}$ converges for $\Re(s)>2k-1$ but I don't know how I can choose $\epsilon.$ Could you Could you give me a hint ? Many thanks – Med Mar 20 '16 at 13:43
  • Can you see that you need $\overbrace{2k-2+4\epsilon}^{\text{exponent of $n$ in the numerator}}-\overbrace{\ \ \ \ \mathrm{Re}(s)\ \ \ \ }^{\text{exponent of $n$ in the denominator}}\lt-1$? – robjohn Mar 20 '16 at 13:57
  • Yes this imply convergence for $\mathrm{Re}(s)>2k-1+4\epsilon$ so we must choose $\epsilon=0$ but the inequality $a(n)^2\ll_{\epsilon}n^{2k-2+4\epsilon}$ satisfied just for $\epsilon>0.$ – Med Mar 20 '16 at 14:03
  • Why must you choose $\epsilon=0$? All you need is that $4\epsilon\lt \mathrm{Re}(s)-(2k-1)$. – robjohn Mar 20 '16 at 14:05
  • Thank you so much for the explanation. – Med Mar 20 '16 at 14:09