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

Question

3111 views around the world

You can reuse this answer
Creative Commons License

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#.