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

Question

Cameron Alexander · Accepted Answer

#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#.

Christopher Agresta · Answer

undefined To find the positive values of $ p $ for which the series $ \sum_{n=2}^{\infty} n(1+n^2)^p $ converges, we can use the limit comparison test or the ratio test.

First, let's consider the ratio test. We evaluate the limit:

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$

where $ a_n = n(1+n^2)^p $.

$$ \lim_{n \to \infty} \left| \frac{(n+1)(1+(n+1)^2)^p}{n(1+n^2)^p} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{(n+1)(1+n^2+2n+1)^p}{n(1+n^2)^p} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{(n+1)(n^2+2n+2)^p}{n(1+n^2)^p} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{(n+1)(n^2(1+2/n+2/n^2))}{n(1+1/n^2)^p} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{n^3(1+2/n+2/n^2)}{n(1+1/n^2)^p} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{n^3(1+2/n+2/n^2)}{n(1+0)^p} \right| $$

$$ = \lim_{n \to \infty} \left| \frac{n^3(1+2/n+2/n^2)}{n} \right| $$

$$ = \lim_{n \to \infty} \left| n^2 + 2 + 2/n \right| $$

As $ n \to \infty $, $ n^2 $ dominates the expression, and the limit becomes $ \infty $.

Since the limit is greater than 1, the series diverges for all $ p $ when $ p > 0 $. Therefore, there are no positive values of $ p $ for which the series converges.