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On the Sum of Reciprocals of Mersenne Primes

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1. Introduction

Since monk Marine Mersenne studied the primality of ${M}_{n}={2}^{n}-1$ in 1644, Mersenne primes, i.e., ${M}_{n}={2}^{n}-1$ ( ${M}_{n}$ : prime), have been developed by numerous researchers, such as Euler, Lucas, Pervouchine, Cole, and Powers, and in recent years, by GIMPS (Great Internet Mersenne Prime Search).

If ${M}_{n}$ is prime, then n is also prime, because if $n=ab$ , ( $a,b\ge 2$ ), then ${M}_{n}=11\cdots 1$ (ab digits in binary) can be divided by $1\cdots 1$ (a digits in binary). However, the converse is not true, for example, ${M}_{11}=2047=23\times 89$ .

In addition, it is well known that all even perfect numbers (odd perfect numbers are unknown) are generated by ${2}^{n-1}{M}_{n}$ , if and only if ${M}_{n}$ is prime.

The current Mersenne prime numbers are denoted by ${M}_{n}$ , for

$n=2,3,5,7,13,17,19,31,61,89,107,127,521,\cdots $ . The most recent Mersenne prime number is ${M}_{74207281}$ (22338618 digits), which was developed in January 2016.

2. Bounds for the Sum of Reciprocals of Mersenne Primes

We begin by defining the notation. We define

$S\equiv \left\{2,3,5,7,13,17,19,31,61,89,107,127,521,\cdots \right\}$

${i}_{k}$ : k-th number in S.

Theorem 1.

$\underset{k=1}{\overset{N}{{\displaystyle \sum}}}\frac{1}{{M}_{i}{}_{k}}<\underset{{M}_{n}:\text{primes}}{{\displaystyle \sum}}\frac{1}{{M}_{n}}<\underset{k=1}{\overset{N}{{\displaystyle \sum}}}\frac{1}{{M}_{i}{}_{k}}+{2}^{2-{i}_{N+1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\forall {i}_{N}\in S$

Proof.

$\begin{array}{c}\underset{{M}_{n}:\text{primes}}{{\displaystyle \sum}}\frac{1}{{M}_{n}}=\underset{i\in S}{{\displaystyle \sum}}\frac{1}{{M}_{i}}=\underset{k=1}{\overset{N}{{\displaystyle \sum}}}\frac{1}{{M}_{i}{}_{k}}+\underset{k=N+1}{\overset{\infty}{{\displaystyle \sum}}}\frac{1}{{M}_{i}{}_{k}}<\underset{k=1}{\overset{N}{{\displaystyle \sum}}}\frac{1}{{M}_{i}{}_{k}}+\underset{k=N+1}{\overset{\infty}{{\displaystyle \sum}}}\frac{1}{{2}^{{i}_{k}}-1}\\ <\underset{k=1}{\overset{N}{{\displaystyle \sum}}}\frac{1}{{M}_{i}{}_{k}}+\underset{k=N+1}{\overset{\infty}{{\displaystyle \sum}}}\frac{1}{{2}^{{i}_{k}-1}}=\underset{k=1}{\overset{N}{{\displaystyle \sum}}}\frac{1}{{M}_{i}{}_{k}}+{2}^{2-{i}_{N+1}}\end{array}$

We can effectively calculate $\underset{{M}_{n}:\text{primes}}{{\displaystyle \sum}}\frac{1}{{M}_{n}}$ , as ${2}^{2-{i}_{N+1}}$ rapidly converges to 0.

For example, if we consider $N=8$ , we obtain

$0.51645417894078856489\cdots <{\displaystyle {\sum}_{{M}_{n}:\text{primes}}\frac{1}{{M}_{n}}}<0.51645417894078856663\cdots $ ,

which provides the value of $\sum}_{{M}_{n}:\text{primes}}\frac{1}{{M}_{n}$ up to 17 decimal digits. If we con-

sider $N=12$ , we can precisely calculate the sum of reciprocals of Mersenne primes up to 156 decimal digits, which is given by

0.516454178940788565330487342971522858815968553415419701441931

065273568701440212723499154883293666215374032432110836569575

419140470924868317486037285294641634・・・

3. Comments

According to the Goldbach-Euler theorem [1] ,

$\underset{q\in \mathbb{P}}{{\displaystyle \sum}}\frac{1}{q-1}=\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\cdots =1,$

where $\mathbb{P}\equiv \left\{{p}^{i}|p\text{\hspace{0.17em}}\text{isaprimenumber}\text{\hspace{0.17em}}\ge 2\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}i\ge 2\right\}$ is the set of perfect powers of

prime numbers.

Theorem 2. The sum of reciprocals of Mersenne prime numbers is larger than that of $q-1$ where $q\in \mathbb{P}\text{\hspace{0.17em}}|p\ge 3$ , namely,

$\underset{{M}_{n}:\text{primes}}{{\displaystyle \sum}}\frac{1}{{M}_{n}}>\underset{\left\{q\in \mathbb{P}\text{\hspace{0.17em}}\mid p\ge 3\right\}}{{\displaystyle \sum}}\frac{1}{q-1}$

Proof. It holds that

$\frac{1}{{M}_{2}}+\frac{1}{{M}_{3}}+\frac{1}{{M}_{5}}\approx 0.50844854\cdots <\underset{{M}_{n}:\text{primes}}{{\displaystyle \sum}}\frac{1}{{M}_{n}}.$

Considering that ${M}_{n}:\text{primes}\in {2}^{i}-1\text{\hspace{0.17em}}\left(i\ge 2\right)$ , it follows from Goldbach-Euler theorem that

$\underset{{M}_{n}:\text{primes}}{{\displaystyle \sum}}\frac{1}{{M}_{n}}+\underset{\left\{q\in \mathbb{P}\text{\hspace{0.17em}}\mid p\ge 3\right\}}{{\displaystyle \sum}}\frac{1}{q-1}<\underset{i\ge 3}{{\displaystyle \sum}}\frac{1}{{2}^{i}-1}+\underset{\left\{q\in \mathbb{P}\text{\hspace{0.17em}}\mid p\ge 3\right\}}{{\displaystyle \sum}}\frac{1}{q-1}=1.$

Hence,

Figure 1. Relationship between ${i}_{n}$ and $\mathrm{log}{M}_{n}$ .

$\underset{\left\{q\in \mathbb{P}\text{\hspace{0.17em}}\mid p\ge 3\right\}}{{\displaystyle \sum}}\frac{1}{q-1}<\frac{1}{2}<\underset{{M}_{n}:\text{primes}}{{\displaystyle \sum}}\frac{1}{{M}_{n}},$

since $\frac{1}{2}<\underset{{M}_{n}:\text{primes}}{{\displaystyle \sum}}\frac{1}{{M}_{n}}$ .

We should note that the sum of reciprocals of prime numbers appears to converge numerically; however, it is infinite, which is proved in, e.g., Hardy and Wright [2] . Therefore, Mersenne primes are considerably sparse subsequences of prime numbers.

In the case of twin primes, the value of the sum of reciprocals of twin primes is shown to be bounded above by Brun [3] and is estimated as

$\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{5}+\frac{1}{7}\right)+\left(\frac{1}{11}+\frac{1}{13}\right)+\left(\frac{1}{17}+\frac{1}{19}\right)+\cdots \approx 1.9021605823\pm 8\times {10}^{-10},$

which is known as Brun’s constant (however, it is an estimation). Even though the problem that whether twin primes are infinite is still unsolved, Zhang [4] presented an important result, which states that a constant $<7\times {10}^{7}$ exists between two successive primes that are infinite. If we can lower the upper bound by 4, the twin prime conjecture will be solved. The Polymath8 project has reduced the upper bound by four digits.

In addition, the problem that whether Mersenne primes are infinite is still unresolved. Figure 1 shows the relationship between ${i}_{n}$ and $\mathrm{log}{M}_{n}$ .

Denote $m\left(x\right)$ the number of Mersenne primes that do not exceed x. Then, as $m\left(x\right)={i}_{n}$ for $x={M}_{n}$ , it seems from Figure 1 that

$m\left(x\right)\propto \mathrm{log}x+c,$

numerically, where c is a constant. In other words, the number of Mersenne primes tends to increase by a constant per digit.

Conflicts of Interest

The authors declare no conflicts of interest.

Cite this paper

*American Journal of Computational Mathematics*,

**7**, 145-148. doi: 10.4236/ajcm.2017.72012.

[1] |
Bibiloni, L., Viader, P. and Paradís, J. (2006) On a Series of Goldbach and Euler. The American Mathematical Monthly, 113, 206-220. https://doi.org/10.2307/27641889 |

[2] | Hardy, G.H. and Wright, E.H. (1979) An Introduction to the Theory of Number. 5th Edition, Oxford University Press, New York. |

[3] |
Brun, V. (1919) La série 1/5+1/7+1/11+1/13+1/17+1/19+1/29+1/31+1/41+1/43+1/59+1/61+··· les dénominateurs sont nombres premiers jumeaux est convergente où finie. Bulletin des Sciences Mathématiques, 43, 124-128. (In French) |

[4] |
Zhang, Y. (2014) Bounded Gaps between Primes. Annals of Mathematics, 179, 1121-1174. https://doi.org/10.4007/annals.2014.179.3.7 |

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