#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#.