Factor sums

$2^{31}-1$ is a Mersenne prime. We can convert the expression to:

$2 \times 2 \times 2 \times 2 \times ...$ (30 of these) $\times 2^{31} -1$

The factors would be:

• $2^{0}$ , $2^{1}$ , ... $2^{31}$ ,
• $2^{0} \times (2^{31} -1)$ , $2^{1} \times (2^{31} -1)$ , ... $2^{31} \times (2^{31} -1)$

Start with the first half of factors. Because of Zeno's Paradox, we know that $\frac{1}{2^{0}} \times \frac{1}{2^{1}} \times \frac{1}{2^{2}} \times ...$ Is going to get VERY close to 2, so close in fact, it may as well be 2. (We can round our answer to the nearest hundredth, so this is completely ok) (It's not infinitely close to 2, but it's just so close that we can assume that when rounded, it will be 2). When we introduce the gigantic prime, the factors become so huge (To become VERY small, when taking the reciprocal of course), that our final answer 2 won't be even remotely changed past the hundredth's place. Therefore, it must be $\boxed{2}$ .

It is known $2^{31}-1$ is a Mersenne prime. Let $p = 2^{31}-1$ . We then have our sum

$\sum_{d|2^{30}(2^{31}-1)} \frac{1}{d} = \sum_{k=0}^{31} \frac{1}{2^{k}} + \sum_{k=0}^{31} \frac{1}{2^{k}p} = (1+\frac{1}{p}) \sum_{k=0}^{31} \frac{1}{2^{k}} \approx 1 \times \sum_{k=0}^{\infty} \frac{1}{2^{k}} = 2$

since $p >> 0$ .