Homework Soursesummation n1infinity 1nlnn12Solutionn n nlogn n for n 3 or
(summation) n=1,[infinity]. 1/(n(ln(n+1))^2
Solution
n = n
nlog(n) >= n for n>= 3
or
n2 [log(n)]2 > n2
1/n2 [log(n)]2 < 1/n2
take summation both sides :
1/n2 [log(n)]2 < 1/n2
as 1/n2 is convergent !!! using p series test !!!
![(summation) n=1,[infinity]. 1/(n(ln(n+1))^2Solutionn = n nlog(n) >= n for n>= 3 or n2 [log(n)]2 > n2 1/n2 [log(n)]2 < 1/n2 take summation both sides](/WebImages/16/summation-n1infinity-1nlnn12solutionn-n-nlogn-n-for-n-3-or-1029280-1761533245-0.webp "(summation) n=1,[infinity]. 1/(n(ln(n+1))^2Solutionn = n nlog(n) >= n for n>= 3 or n2 [log(n)]2 > n2 1/n2 [log(n)]2 < 1/n2 take summation both sides")
