1 1 vote Let $T(n) = T(n-1) + \frac{1}{n} , T(1) = 1 ;$ then $T(n) = ? $ $O(n^{2})$ $O(logn)$ $O(nlogn)$ $O(n^{2}logn)$ Combinatory discrete-mathematics recurrence-relation relations + – Lakshman Bhaiya 2.5k views answer comment Share Follow Print See all 3 Comments 3 3 Comments reply pilluverma123 commented Oct 5, 2018 reply Follow flag If we use H.P. sum formula it will come as log(something). So by intuitively we can say option B is correct 0 0 replyShare Lakshman Bhaiya commented Oct 5, 2018 reply Follow flag yeah thanks 0 0 replyShare air1ankit commented Oct 5, 2018 reply Follow flag O(logn) 0 0 replyShare Please log in or register to add a comment.
2 2 votes ...... this is the way to dealing with such type of question air1ankit answered Oct 5, 2018 air1ankit comment Share Follow See all 5 Comments 5 5 Comments reply Show 2 previous comments air1ankit commented Oct 8, 2018 reply Follow flag O(logn) ....And I don't think that any one can explain more for this ...This is more than enough by the way wich part have you facing problem ???? 0 0 replyShare Lakshman Bhaiya commented Oct 8, 2018 reply Follow flag Ok, your method is good, but How you take the condition where $n>1$ given in the question, you take $k =n-1$,right?? if $n>=1$ given the question, what we take $k =?$ if $n>=0$ given the question, what we take $k =?$ 0 0 replyShare air1ankit commented Oct 11, 2018 reply Follow flag I also have same doubt . Unable to explain sorry bro ..! 0 0 replyShare Please log in or register to add a comment.
0 0 votes T(n)=1/1+1/2+1/3+1/4+.......+1/n T(n)= O(logn) Raghava45 answered May 14, 2019 Raghava45 comment Share Follow 0 reply Please log in or register to add a comment.
