How do you find the limit of #s(n)=64/n^3[(n(n+1)(2n+1))/6]# as #n->oo#? Calculus Tests of Convergence / Divergence Partial Sums of Infinite Series 1 Answer Shwetank Mauria Dec 10, 2017 #lim_(n->oo)s(n)=lim_(n->oo)64/n^3[(n(n+1)(2n+1))/6]=21 1/3# Explanation: #lim_(n->oo)s(n)=lim_(n->oo)64/n^3[(n(n+1)(2n+1))/6]# = #lim_(n->oo)64/6[n/n*(n+1)/n*(2n+1)/n]# = #lim_(n->oo)64/6[1*(1+1/n)*(2+1/n)]# = #64/6*1*1*2# = #64/3=21 1/3# Answer link Related questions How do you find the n-th partial sum of an infinite series? How do you find the n-th partial sum of a geometric series? How do you find the 5-th partial sum of the infinite series #sum_(n=1)^oo1/(n(n+2)# ? How do you find the 10-th partial sum of the infinite series #sum_(n=1)^oo(0.6)^(n-1)# ? How do you find the 6-th partial sum of the infinite series #sum_(n=1)^oo1/n# ? How do you find the 4-th partial sum of the infinite series #sum_(n=1)^oo(1/sqrt(n)-1/sqrt(n+1))# ? How do you find the 4-th partial sum of the infinite series #sum_(n=1)^oo(3/2)^n# ? How do you find the 5-th partial sum of the infinite series #sum_(n=1)^ooln((n+1)/n)# ? How do you find the sum of the series #1+ln2+(((ln2)^2)/(2!))+...+(((ln2)^2)/(n!))+...#? How do you find partial sums of infinite series? See all questions in Partial Sums of Infinite Series Impact of this question 4672 views around the world You can reuse this answer Creative Commons License