3.1.9 problem 10

(a)

Let $P\left(n\right)=\frac{k}{n}$ for $n\in N$. By principles of probability, ${\sum }_{n\in N}P\left(n\right)$ must equal $1$.

$$\sum _{n\in N}P\left(n\right)=k\sum _{n\in N}\frac{1}{n}.$$

The sum on the right side of the equality is a divergent, harmonic series. Hence. the aforementioned principle of probably is violated. Contradiction.

(b)

${\sum }_{n\in N}\frac{1}{n^2}=\frac{{\pi }^2}{6}$. Thus, letting $k$ equal $\frac{6}{{\pi }^2}$, the principle of probability is satisfied.