0 votes 0 votes How is Σ (1/k * log k) = O( n log n) for k=1 to n? Algorithms algorithms time-complexity asymptotic-notation + – Nidhi Budhraja asked Sep 7, 2018 Nidhi Budhraja 339 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply BharathiCH commented Sep 7, 2018 i edited by BharathiCH Sep 7, 2018 reply Follow Share ∑(1/k)*logk=(log1)/1 +(log2)/2 + (log3)/3 + .... + (logn)/n =((log1)(2*3*4*..*n)+(log2)(1*3*4*5...*n)+....+(logn)(1*2*3*...*n-1)) / (1*2*..*n) =((log1)(2*3*4*..*n)+(log2)(1*3*4*5...*n)+....+(logn)(1*2*3*...*n-1)) / n! ≤ log (n*n*n*...*n) ≊ log(n^n) = O(nlogn) 1 votes 1 votes Nidhi Budhraja commented Sep 7, 2018 reply Follow Share Thanks 0 votes 0 votes nephron commented Oct 31, 2018 reply Follow Share BharathiCH Nidhi Budhraja I am getting O(log2n) . What is the answer given??? 0 votes 0 votes Please log in or register to add a comment.
