Question

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

Answer 1

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.