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 · 2016-12-01T18:54:12

#sum_n n(1+n^2)^p# is divergent for all positive values of #p#.

#a_n=n(1+n^2)^p#

As #a_n > 0#, a necessary condition for #sum a_n# to converge is that:

#lim_n a_n = 0#

But as #(1+n^2) > 1#, for any #p>0#

#lim_n n(1+n^2)^p = oo#

and the series is divergent.

We can use the integral test of convergence to find other values of #p# for which the series converges, finding a function #f(x)# such that:

#f(n)=a_n#

Certainly:

#f(x) = x(1+x^2)^p#

fits the purpose, so the series:

#sum n(1+n^2)^p#

will converge if:

#F(x) = int_1^(oo) x(1+x^2)^pdx#

also converges.

Substitute #x^2=t#:

#int_1^(oo) x(1+x^2)^pdx = 1/2int_1^(oo) (1+t)^pdt =(1+x^2)^(p+1)/(2(p+1))|_(x=1)^(x->oo)#

In #t=1# the value is always finite except for #p=-1# and equal to:

#F(1) = 2^p/(p+1)#

Let's look at:

#lim_(x->+oo) (1+x^2)^(p+1)#

This limit converges only for #(p+1)<0# or #p < -1#.

So the series in convergent only for #p < -1#.