This is from DeGroot's "PROBABILITY and STATISTICS"(Second edition)(Cf. pages 87 to 92).I am rewriting the relevant stuff.
Let $r$ be a positive integer and $r\leq n$. Define$$P_r=\frac{1}{n}+\frac{r-1}{n}\bigg(\frac{1}{r}+\frac{1}{r+1}+\dots+\frac{1}{n-1}\bigg).$$And thus,$$P_{r+1}-P_r=\frac{1}{n}\bigg(\frac{1}{r}+\frac{1}{r+1}+\dots+\frac{1}{n-1}-1\bigg).\tag{1}$$It can be seen from Equation $(1)$ that thedifference $P_{r+1}-P_r$ is a decreasing function of $r$. As long as the difference is positive,$P_{r+1}$ will be larger than $P_r$.However, as soon as this difference becomes negative for some value of $r$,then $P_{r+1}$ will be smaller than $P_r$ and it will continue to decrease for all larger values of $r$.Hence, the value of $r$ for which $P_r$ is a maximum will be the smallest value of $r$ for which $(P_{r+1}-P_r)\leq 0$.It follows from Equation $(1)$that we must find the smallest value of $r$ such that $$\bigg(\frac{1}{r}+\frac{1}{r+1}+\dots+\frac{1}{n-1}\bigg)\leq1.\tag2$$ We shall call this value $r^*$. Since $r^*$ is the smallest value of $r$ for which Relation $(2)$ is satisfied,it will be approximately true for large values of $n$ that $$\bigg(\frac{1}{r^*}+\frac{1}{r^*+1}+\dots+\frac{1}{n-1}\bigg)\approx1.\tag3$$
Now, I don't understand these two things:
$(1)$ Why the value of $r$ for which $P_r$ is a maximum will be the smallest value of $r$ for which $(P_{r+1}-P_r)\leq 0$?
$(2)$ How Approximation $(3)$ holds?
Please help.