How do you find the positive values of p for which $Sigma n/(1+n^2)^p$ from $[2,oo)$ converges?

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Answer 1 · 2017-07-20T21:42:12

The series:

$sum_(n=2)^oo n/(1+n^2)^p$

is convergent for $p > 1$

Consider the series:

$sum_(n=1)^oo 1/n^(2p-1)$

Based on the $p$ -series test, this series is convergent for $2p-1 > 1$ , that is for $p> 1$ .

Using this series for the limit comparison test:

$lim_(n->oo) (n/(1+n^2)^p)/(1/n^(2p-1)) = lim_(n->oo) n^(2p)/(1+n^2)^p$

$lim_(n->oo) (n/(1+n^2)^p)/(1/n^(2p-1)) = lim_(n->oo) (n^2/(1+n^2))^p =1^p = 1$

which means that the two series have the same character. Then also the first series is convergent for $p>1$ .

How do you find the positive values of p for which Sigma n/(1+n^2)^pSigma n/(1+n^2)^p from [2,oo)[2,oo) converges?