About a step in the analysis of Quicksort by Sedgewick and Wayne
Let $\rho_N = C_N/(N+1)$. The next to last equation shows that
$$
\rho_N = \rho_{N-1} + \frac{2}{N+1}.
$$
Therefore
$$
\rho_N = \frac{2}{N+1} + \cdots + \frac{2}{2+1} + \rho_1.
$$
Multiplying by $N+1$,
$$
C_N = (N+1) \left(\frac{2}{3} + \cdots + \frac{2}{N+1} + \rho_1\right).
$$
Since $\rho_1$ is a constant whereas $2/3 + \cdots + 2/(N+1) \to \infty$, we deduce that
$$
C_N \sim (N+1) \left(\frac{2}{3} + \cdots + \frac{2}{N+1}\right).
$$
We can estimate this expression in many ways, for example by approximating the series by an integral (this is what the authors suggest when they mention the area under the curve $2/x$). Or we can recognize that it is roughly equal to $2H_{N+1}$, twice the $(N+1)$st harmonic number, and so using the well-known estimate $H_N \sim \log N$,
$$
C_N \sim 2N\log N.
$$

Finally, regarding the relevance to Quicksort, my guess is that $C_N$ is the average number of comparisons performed on a random array of length $N$; but you should be able to tell by reading the book.