Use comparison test to determine convergence of the following series?

#sum_(n-1)^oo(n+1)/n^2#

1 Answer
Mar 27, 2018

#sum_(n-1)^oo(n+1)/n^2# is divergence

Explanation:

Let #a_n# = #(n+1)/n^2# and #b_n# = #1/n# are sequence of positive real numbers.

such that

#L# = #lim_(n->+∞)((n+1)/n^2)/(1/n)# = #lim_(n->+∞)((n+1)/n)# = #1#

Because

#sum_(n-1)^oo(1/n)# is divergence,

so, by using limit ratio test (comparison test),

we said that #sum_(n-1)^oo(n+1)/n^2# is divergence.