How do you test for convergence for #sum(5^k+k)/(k!+3)# from k=1 to infinity? Calculus Tests of Convergence / Divergence Direct Comparison Test for Convergence of an Infinite Series 1 Answer Cesareo R. Jan 29, 2018 See below. Explanation: #sum_(k=1)^oo(5^k+k)/(k!+3) < sum_(k=1)^oo(5^k)/(k!) + sum_(k=1)^oo(k)/(k!) le e^5+e# so it converges. Answer link Related questions What is the Direct Comparison Test for Convergence of an Infinite Series? How do you use the direct comparison test for infinite series? How do you use the direct comparison test for improper integrals? How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo5/(2n^2+4n+3)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooln(n)/n# ? How do you use the direct Comparison test on the infinite series #sum_(n=2)^oon^3/(n^4-1)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo9^n/(3+10^n)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo(1+sin(n))/(5^n)# ? How do you use the direct Comparison test on the infinite series #sum_(n=1)^ooarctan(n)/(n^1.2)# ? How do you use basic comparison test to determine whether the given series converges or diverges... See all questions in Direct Comparison Test for Convergence of an Infinite Series Impact of this question 3204 views around the world You can reuse this answer Creative Commons License