How do you use the comparison test (or the limit comparison test) for #(1+sin(x))/10^x#? Calculus Tests of Convergence / Divergence Limit Comparison Test for Convergence of an Infinite Series 1 Answer Jim H Oct 15, 2015 For the series #sum_(i=1)^oo (1+sin(n))/10^n# use #1+sinn <= 2#. Explanation: So #(1+sin(n))/10^n <= 2/10^n# #sum_(i=1)^oo 2/10^n# is a geometric series with #r = 1/10# so it converges. 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 determine whether #1/(n!)# convergence or divergence with direct comparison test? 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 4868 views around the world You can reuse this answer Creative Commons License