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We know the elementary fact that if the partial sums $ \sum_{n\leq X} a_n $ are bounded, say by $ C$, then the series $ \sum_{n\geq 1} a_n n^{-s} $ converges for $s >0$.

My question then is, is there a simple upper bound in terms of $X$ and $C$ of the tail series $\sum_{n\geq X} a_n n^{-s} $ for a complex number $s $ with $\Re s>0$ ? The upper bound should tend to $0$ as $X\to \infty$.

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Take $Y>X>0$ two integers and $s = a + ib$. Denote $S_X = \sum_{n=1}^X a_n n^{-s}$. Then $|S_Y - S_X| = | \sum_{n=X+1}^Y a_n n^{-a} n^{-ib}| \leq \sum_{n=X+1}^Y |a_n n^{-a}| \leq C (X+1)^{-a}$. You get convergence using the Cauchy caracterisation and an upper bound $| \sum_{n \geq X} a_n n^{-s} | \leq C X^{-\Re(s)}$.

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  • $\begingroup$ Thanks. But there is an error; I got the idea nevertheless. $\endgroup$
    – U Ser
    Apr 14, 2017 at 12:41

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