Task B-4.8. Determine the sum of the reciprocals of all positive divisors of the number $n$, including 1 and $n$, if $n=2^{p-1}\left(2^{p}-1\right)$, and $2^{p}-1$ is a prime number.
Solution. The divisors of the number $n$ are the numbers

$$
1,2,2^{2}, 2^{3}, \ldots, 2^{p-1}, 2^{p}-1,2 \cdot\left(2^{p}-1\right), 2^{2} \cdot\left(2^{p}-1\right), \ldots, 2^{p-1} \cdot\left(2^{p}-1\right)
$$

The sum of their reciprocals is

$$
S=1+\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{p-1}}+\frac{1}{2^{p}-1}\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\ldots+\frac{1}{2^{p-1}}\right)
$$

We observe a geometric sequence of $p$ terms with the first term 1 and the common ratio $\frac{1}{2}$. Then,

$$
S=\frac{1-\left(\frac{1}{2}\right)^{p}}{\frac{1}{2}} \cdot\left(1+\frac{1}{2^{p}-1}\right)=2\left(1-2^{-p}\right)\left(\frac{2^{p}}{2^{p}-1}\right)=2\left(\frac{2^{p}-1}{2^{p}-1}\right)=2
$$

(4 points)