How do you determine whether #1/(n!)# convergence or divergence with direct comparison test? Calculus Tests of Convergence / Divergence Limit Comparison Test for Convergence of an Infinite Series 1 Answer Jim H Apr 14, 2015 Compare it to #1/n^2#. For #n>=4#, we have #1/(n!) < 1/n^2# Answer link Related questions How do you use the limit comparison test on the series #sum_(n=1)^oon/(2n^3+1)# ? How do you use the limit comparison test on the series #sum_(n=1)^oo(n+1)/(n*sqrt(n))# ? How do you use the limit comparison test on the series #sum_(n=2)^oosqrt(n)/(n-1)# ? How do you use the limit comparison test on the series #sum_(n=1)^oo(n^2-5n)/(n^3+n+1)# ? How do you use the limit comparison test on the series #sum_(n=1)^oo1/sqrt(n^3+1)# ? What is the Limit Comparison Test? How do I use the Limit Comparison Test on the series #sum_(n=1)^oosin(1/n)# ? How do I know when to use limit comparison test vs the direct comparison test? How do you use the comparison test (or the limit comparison test) for #(1+sin(x))/10^x#? How do you do the limit comparison test for this problem #sqrt ( (n+1)/ (n^2+2))# as n goes to infinity? See all questions in Limit Comparison Test for Convergence of an Infinite Series Impact of this question 20887 views around the world You can reuse this answer Creative Commons License