Using the integral test, how do you show whether #sum 1/((2n+1)^2)# diverges or converges from n=1 to infinity?

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Answer 1 · 2018-05-22T13:35:38

The series converges.

First observe that

#(2n+1)^2>=4n^2#

#1/(2n+1)^2<=1/(4n^2)#

since #n>=1#

#sum1/(4n^2)>=sum1/(2n+1)^2#

Using the integral test, we have

#int_1^oo1/(4x^2)dx=1/4[-1/x]_1^oo#

#=1/4(0+1)#

#=1/4#

So,

#sum1/(4n^2)# converges and hence, so does

#sum1/(2n+1)^2#