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 5688 views around the world You can reuse this answer Creative Commons License