How do you find the 6-th partial sum of the infinite series #sum_(n=1)^oo1/n# ? Calculus Tests of Convergence / Divergence Partial Sums of Infinite Series 1 Answer VNVDVI Apr 21, 2018 #S_6=49/20# Explanation: Let's get a formula for the #kth# partial sum: #S_k=sum_(n=1)^k1/n=1+1/2+1/3+...+1/k# The #6th# partial sum is then obtained as follows: #S_6=sum_(n=1)^6 1/n=1+1/2+1/3+1/4+1/5+1/6=49/20# 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 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? How do you write a sum in expanded form? See all questions in Partial Sums of Infinite Series Impact of this question 5364 views around the world You can reuse this answer Creative Commons License