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I'd like to get an asymptotic formula for $\sum_{\substack{n\leq x\\ n\text{ square free}}}\frac{1}{n^l}$ for $l>0$.

We know that $\sum_{\substack{n\leq x\\ n\text{ square free}}}1= cx+O(\sqrt{x})$.

Suppose $l>1$, if I use partial summation for $\sum_{\substack{n\leq x\\ n\text{ square free}}}\frac{1}{n^l}$, then I get the error term is larger than main term (since the main term is $x^{1-l}$ and the error term is $O(1)$). Is there a way to get an asymptotic formula for $\sum_{\substack{n\leq x\\ n\text{ square free}}}\frac{1}{n^l}$?

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  • $\begingroup$ What’s $\,\,\, n\ square\ free$ ?. $\endgroup$
    – Felix Marin
    Commented Jul 4 at 16:59
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    $\begingroup$ @FelixMarin It means the summation is over square-free integers $n$. An integer is square-free if its prime factorization has exactly one factor for each prime that appears in it. $\endgroup$
    – Nick
    Commented Jul 4 at 17:15
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    $\begingroup$ Apart from the "square free" condition, you may want to check Generalized harmonic numbers, e.g. here: en.wikipedia.org/wiki/Harmonic_number $\endgroup$
    – Andreas
    Commented Jul 4 at 18:09
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    $\begingroup$ Well when x is taken to infinity is equal too $\frac{\zeta (\lambda)}{\zeta (2\lambda)}$ as the square free condition is just the absolute value of the mobius function. Im unsure how to find the exact asmytotics for partial expansions. $\endgroup$
    – Aidan R.S.
    Commented Jul 4 at 19:32
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    $\begingroup$ If you only care about the main term then a quick calculation using partial summation gives $(c-\frac{l}{l+1})x^{1-l}$ for $l<1$, and for $l>1$ the main term is $\zeta(l)/\zeta(2l)$ as mentioned by @AidanR.S. $\endgroup$
    – Nah
    Commented Jul 6 at 2:17

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