How to calculate the limit of #(ln2/2+ln3/3+...+lnn/n)/ln^2n# when #x->∞# ?

1 Answer
Jan 30, 2018

0

Explanation:

If we just look at the last summand of the fraction and divide it by the denominator #ln^2n#, we get
#(lnn)/(n*ln^2n)=(cancel(lnn))/(n*lnn*cancel(lnn))=1/(n*lnn)#
If n goes towards infinity, this fraction will go towards 0, because #n*lnn# is divergent.
Because all the other fractions are smaller than #lnn/n# and #ln^2n#, which is for large values larger than 1, is going to divide them, they will still be smaller than #lnn/n#. Therefore, the limit is equal to 0.