### Euler-Mascheroni constant

In this post we're going to take a look at a mysterious mathematical constant, called the Euler–Mascheroni constant, and its fascinating role in harmonic and prime numbers.

`lim_{n to \infty} \frac{\sigma(n)}{\H_n + \ln(\H_n)*\exp(\H_n)} = 1`

suggesting that the Riemann hypothesis is true.

### # Is `gamma` transcendental?

This constant, although it has a fairly simple definition, it is currently not known whether it is rational or irrational, but it is widely believed by mathematicians to be transcendental, which also implies that it is irrational.

It is usually defined as:

`\gamma = \lim_{n to \infty}(\H_n - \ln n)`

where `\H_n` is the `n`

*harmonic number, which is defined as:*^{th}
`\H_n = \sum_{k=1}^(n)\frac{1}{k}`

### # "Proving" that `\gamma` is transcendental

There exists a

*proof*that `\gamma` is transcendental, but the*proof*is very subtle:- By Lindemann–Weierstrass theorem, the natural logarithm of any positive algebraic number other than 1 is a transcendental number.
- The `n`
harmonic number is rational.^{th}

As all rational numbers are algebraic numbers, the definition of `\gamma` reduces to:

**algebraic - transcendental = transcendental**
which concludes the

*proof*that `\gamma` is transcendental.
The second statement of the

*proof*is the most critical part, which implies the following:
`\lim_{n to \infty}\sum_{k=1}^(n)\frac{1}{k} = \frac{a}{b}`

for some integer `a` and `b`.

Although it has been shown that the harmonic series diverges, the

*proof*requires that the `n`*harmonic number will always be a rational number, no matter how large `n` becomes.*^{th}
If we consider the concept of infinity as a process that never ends, then the above summation, based on the proof that the harmonic series is divergent, implies that `a/b` become larger and larger as the process continues, but it will

**never**reach a point where `a/b = \infty`.### # The role of `\gamma` in prime numbers

The most fascinating role of the `\gamma` constant is its role in prime numbers:

`\lim_{n to \infty}\ln n \prod_{p}^(n)(1-\frac{1}{p}) = \exp(-\gamma)`

Formulating the result a little bit differently, letting `z` to be defined as:

`z = \lim_{n to \infty}\prod_{p}^(n)p`

we will have:

`\frac{\phi(z)}{z} = \lim_{n to \infty}\frac{\exp(-\gamma)}{\ln n}`

where `\phi(z)` is Euler's totient function.

The same result can be illustrated in yet another way, using the positive value of the `gamma` constant:

`\lim_{n to \infty}\frac{1}{\ln p_n}\prod_{k=1}^(n)\frac{\p_k}{\p_k - 1} = \exp(\gamma)`

where `\p_n` is the `n`

*prime number.*^{th}### # Harmonic numbers

Interestingly enough, any harmonic number can be represented in terms of the `gamma` constant:

`\H_n = \psi(n+1) + \gamma`

Harmonic numbers play an important role in number theory; as proven in 2002 by Jeffrey Lagarias, the Riemann hypothesis is equivalent to the following statement:

`\sigma(n) <= \H_n + \ln(\H_n)*\exp(\H_n)`

where `\sigma(n)` is the sum of the positive divisors of `n`.

However, proving this is not that as easy as it may seem to be at the first glance. A potentially good strategy would be to restrict `n` only to highly composite numbers (A002182) or highly abundent numbers (A002093) and prove that `\sigma(n)` does not exceed `\H_n + \ln(\H_n)*\exp(\H_n)`, which would automatically prove it for all the other numbers.

For highly composite numbers `n`, we can (empirically) observe that:

suggesting that the Riemann hypothesis is true.

We can illustrate this with the largest highly composite number currently known (?):

`a = P(42681)*P(151)*P(29)*P(11)*P(7)*P(5)*P(4)*P(3)^2*P(2)^4*P(1)^9`

where `P(n) = \prod_{k=1}^(n)p_k`

leading to:

`\frac{\sigma(a)}{\H_a + \ln(\H_a)*\exp(\H_a)} ~~ 0.999910909330009471428528222805...`

In 1913, Grönwall showed that the asymptotic growth rate of the sigma function can be expressed by:

`\lim_{n to \infty}\frac{\sigma(n)}{\n \ln\ln n} = \exp(\gamma)`

where

From this we can conclude the following result:

where, for `n = \Gamma(10^7+1)`, we have:

`\frac{\exp(\gamma) * n \ln \ln n}{\H_n + \ln(\H_n) * \exp(\H_n)} ~~ 0.999999999797279018107926...`

illustrating the convegence towards `1`.

Slightly less elegantly, we can reformulate it as (see Wolfram|Alpha):

`\lim_{n to \infty}\frac{\exp(\gamma) * n \ln \ln n}{\exp(\gamma) * n * \ln(\gamma + \ln n) + \gamma + \ln n} = 1`

For example:

`zeta((10^6 + 1)/10^6) ~~ 10^6 + 0.577215737717373499101298...`

This gives us the limit:

`\lim_{n to \infty}[zeta((n+1) / n)-n] = \gamma`

`z = \prod_{p}^(n)p`

we can extract the approximative value of `n`, using the following formula:

`n ~~ \exp(\frac{z * \exp(-\gamma)}{\phi(z)})`

###

The `\gamma` constant seems to be one of the most important and interesting mathematical constants, despite the fact that we only know relatively few things about it.

See also:

`a = P(42681)*P(151)*P(29)*P(11)*P(7)*P(5)*P(4)*P(3)^2*P(2)^4*P(1)^9`

where `P(n) = \prod_{k=1}^(n)p_k`

leading to:

`\frac{\sigma(a)}{\H_a + \ln(\H_a)*\exp(\H_a)} ~~ 0.999910909330009471428528222805...`

In 1913, Grönwall showed that the asymptotic growth rate of the sigma function can be expressed by:

`\lim_{n to \infty}\frac{\sigma(n)}{\n \ln\ln n} = \exp(\gamma)`

where

*lim*is the limit superior.From this we can conclude the following result:

`lim_{n to \infty}\frac{\exp(\gamma) * n \ln \ln n}{\H_n + \ln(\H_n) * \exp(\H_n)} = 1`

where, for `n = \Gamma(10^7+1)`, we have:

`\frac{\exp(\gamma) * n \ln \ln n}{\H_n + \ln(\H_n) * \exp(\H_n)} ~~ 0.999999999797279018107926...`

illustrating the convegence towards `1`.

Slightly less elegantly, we can reformulate it as (see Wolfram|Alpha):

`\lim_{n to \infty}\frac{\exp(\gamma) * n \ln \ln n}{\exp(\gamma) * n * \ln(\gamma + \ln n) + \gamma + \ln n} = 1`

### # Connection with the zeta function

There is a nice connection between the zeta function and the `gamma` constant when the argument of the zeta function approaches `1`:- `zeta((n+1)/n) ~~ n + \gamma`
- `zeta((n-1)/n) ~~ \gamma - n`

For example:

`zeta((10^6 + 1)/10^6) ~~ 10^6 + 0.577215737717373499101298...`

This gives us the limit:

`\lim_{n to \infty}[zeta((n+1) / n)-n] = \gamma`

### # Primorial and the `\gamma` constant

If we let `z` to be defined as:`z = \prod_{p}^(n)p`

we can extract the approximative value of `n`, using the following formula:

`n ~~ \exp(\frac{z * \exp(-\gamma)}{\phi(z)})`

###

# Conclusion

The `\gamma` constant seems to be one of the most important and interesting mathematical constants, despite the fact that we only know relatively few things about it.See also:

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