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On power free numbers

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Entry 1. Let f(r) be any divergent series of positive terms, q_{r,k} be the r^{th} k-power free number, \zeta(k) be the Riemann Zeta function. Let S_f(r) = \sum_{i=1}^{r} f(i). If  g(x) is Riemann integrable in (0, \infty) then,

\displaystyle{ \sum_{r=1}^{n} f(q_{r,k}) g \Big(\frac{S_f(r)}{\zeta(k)}\Big) \sim \int_{f(1)}^{\frac{S_f(n)}{\zeta(k)}} g(x)dx.}

Corollary 1. As k \rightarrow \infty, \zeta(k) \rightarrow 1. Also every natural number is k-power free when k \rightarrow \infty. Hence the above result reduces to

 \displaystyle{ \sum_{r=1}^{n} f(r) g(S_f(r))\sim \int_{f(1)}^{S_f(n)} g(x)dx}.

Entry 2. Further let q'_{k,n} be the n^{th} k-power containing number and f be any function Riemann integrable in (1,\infty); then,

\displaystyle{ \sum_{k=2}^{\infty}\frac{1}{k} \Big\{\frac{f(q'_{k,1}) + f(q'_{k,2}) + f(q'_{k,3}) +\ldots}{f(q_{k,1}) + f(a_{k,2}) + f(q_{k,3}) +\ldots}\Big\}= 1 - \gamma}.

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