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I have some problem with this parametric series: $ \sum_{i=1}^\infty n^\alpha\left(\frac 1{n^{1/4}}-\sin\left(\frac 1{n^{1/4}}\right)\right) $

which value of $\alpha$ makes the series convergent? And which divergent?

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$\sin x\sim x-x^3/6$ as $x\to0$. Then $$ n^{-1/4}-\sin(n^{-1/4})\sim\frac16\,n^{-3/4}\quad\text{as}\quad n\to\infty. $$ Can yo finish?

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  • $\begingroup$ so i have $\sum_1^\infty \frac16n^{\alpha-3/4}$ so if $\alpha$ is $<-\frac14$ $\endgroup$
    – Marco Toni
    Dec 17, 2015 at 14:56
  • $\begingroup$ Yes, both series have the same behavior by the comparison criterium (they are positive series.) $\endgroup$ Dec 17, 2015 at 15:00
  • $\begingroup$ so i have $\sum_1^\infty \frac16n^{\alpha-3/4}$ so if $\alpha$ is $<-\frac14$ the series is convergent and instead if $\alpha$ is $> -\frac14$ $\endgroup$
    – Marco Toni
    Dec 17, 2015 at 15:02
  • $\begingroup$ is it right? for $\alpha < -\frac14$ is it convergent? $\endgroup$
    – Marco Toni
    Dec 17, 2015 at 15:04
  • $\begingroup$ That is correct. $\endgroup$ Dec 17, 2015 at 15:05

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