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

Answer 1

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#.
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Answer 2

The series (\sum_{n=2}^{\infty} \frac{n}{(1+n^2)^p}) converges if (p > \frac{1}{2}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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