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Euler-Mascheroni Constant Watch

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    I am just curious.

    Is Euler-Mascheroni constant \gamma=\lim_{n\rightarrow\infty}  (\sum_{1\le k \le n}\frac{1}{k}-\ln n) rational or irrational? And, how to prove rationality?
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    It's not known. Now, show that

    \displaystyle \int\limits_{0}^{\infty}\frac{ \ln{x}}{e^x}\;{dx} = -\gamma.
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    Can I do something like this:

    Define x=\lim_{n\rightarrow \infty} b (\gamma-\sum_{1\le k \le b}\frac{1}{k}+\ln b) b! n! (n-1)^{\frac{n(n+1)}{2}}, assuming n\in Z

    If \gamma\in Q then \exists a,b \in Z s.t. \gamma=\frac{a}{b}, b\neq \pm 1

    Then

     x=a(b!n!(n-1)^{\frac{n(n+1)}{2}})-b(\sum_{1\le k \le b}\frac{1}{k}-\ln n )(b! n! (n-1)^{\frac{n(n+1)}{2}}), which, after some evaluation, is an integer.

    Consider the definition of \gamma

     x=b(\sum_{b+1\le k\le n}\frac{1}{k})(b!n!(n-1)^{\frac{n(n+1)}{2}})=b\frac{(h  uge\quad numerator)}{n!}(b!)^2 n! (n-1)^{\frac{n(n+1)}{2}} and it turns out to be an integer as well.

    Can I conclude that the constant in rational?

    Note: Typos may exist.

    EDIT: Do this step by step is probably less confusing.

    Define x=\lim_{n\rightarrow \infty} b (\gamma-\sum_{1\le k \le b}\frac{1}{k}+\ln b)\alpha, assuming n\in Z and \alpha=b! n! (n-1)^{\frac{n(n+1)}{2}}

    If \gamma=\frac{a}{b}

    Then
     x = \lim b(\frac{a}{b}-\sum_{1\le k\le b}\frac{1}{k}+\sum_{1\le k\le \infty}\frac{(-1)^k}{k(n-1)^{k}})\alpha=\lim a \alpha-b(\frac{(huge\quad numerator)}{\alpha})\alpha which is an integer assuming n\in Z.

    Similar procedures but using the definition of the constant which will give a integer as well.

    EDIT2: Typo corrected: "nominator" to "numerator"
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    (Original post by agostino981)
    Can I do something like this:

    Define x=\lim_{n\rightarrow \infty} b (\gamma-\sum_{1\le k \le b}\frac{1}{k}+\ln b) b! n! (n-1)^{\frac{n(n+1)}{2}}, assuming n\in Z

    If \gamma\in Q then \exists a,b \in Z s.t. \gamma=\frac{a}{b}, b\neq \pm 1

    Then

     x=a(b!n!(n-1)^{\frac{n(n+1)}{2}})-b(\sum_{1\le k \le b}\frac{1}{k}-\ln n )(b! n! (n-1)^{\frac{n(n+1)}{2}}), which, after some evaluation, is an integer.

    Consider the definition of \gamma

     x=b(\sum_{b+1\le k\le n}\frac{1}{k})(b!n!(n-1)^{\frac{n(n+1)}{2}})=b\frac{(h  uge\quad nominator)}{n!}(b!)^2 n! (n-1)^{\frac{n(n+1)}{2}} and it turns out to be an integer as well.

    Can I conclude that the constant in rational?

    Note: Typos may exist.
    That didn't make any sense whatsoever.
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    (Original post by L'art pour l'art)
    That didn't make any sense whatsoever.
    I initially wanted to prove for irrationality by contradiction, but it turns out to be something like this. By the way, the above "monster" I wrote used Taylor series of logarithm, just too lazy to write all steps.
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    (Original post by agostino981)
    Define x=\lim_{n\rightarrow \infty} b (\gamma-\sum_{1\le k \le b}\frac{1}{k}+\ln b)\alpha, assuming n\in Z and \alpha=b! n! (n-1)^{\frac{n(n+1)}{2}}
    You do realise that you have \displaystyle n! (n-1)^{\frac{n(n+1)}{2}} as n \to \infty?
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    (Original post by agostino981)

    huge\quad nominator
    Intriguing, though we know what you meant.
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    (Original post by jack.hadamard)
    You do realise that you have \displaystyle n! (n-1)^{\frac{n(n+1)}{2}} as n \to \infty?
    Tends to infinity, but we can't just say it is infinity, because it is never infinity.

    EDIT: I may change the factorial to gamma function to consider values of n other than solely integers.

    EDIT2: If n\in Z, then x\in Z. If n\notin Z, then n \notin Z, I am kind of stuck here, whether should I conclude that \gamma\in Q.

    (Original post by ghostwalker)
    Intriguing, though we know what you meant.
    I am too lazy to introduce a new polynomial or something like that.
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    (Original post by agostino981)
    I am too lazy to introduce a new polynomial or something like that.
    You missed the point. The word is "numerator", not "nominator".
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    (Original post by ghostwalker)
    You missed the point. The word is "numerator", not "nominator".
    lol. Thanks! After all, it is not a word frequently used by me. :P
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    (Original post by agostino981)
    Tends to infinity, but we can't just say it is infinity, because it is never infinity.
    What I am saying is that your limit does not make sense. That way, I can prove pretty much everything to be a rational number.
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    (Original post by jack.hadamard)
    What I am saying is that your limit does not make sense. That way, I can prove pretty much everything to be a rational number.
    You mean it doesn't converge?

    I took the limit because of the definition of the constant.
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    (Original post by agostino981)
    You mean it doesn't converge?

    I took the limit because of the definition of the constant.
    He means that your x\to \infty so you cannot conclude that x\in\mathbb{Z}, because it isn't once we introduce the limit.
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    (Original post by Lord of the Flies)
    He means that your x\to \infty so you cannot conclude that x\in\mathbb{Z}, because it isn't once we introduce the limit (which is required given how you are using that definition of \gamma
    But at first I have already assumed n\in Z as the first case to be considered. I am kind of tired at the moment to do the remaining maths.
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    (Original post by agostino981)
    But at first I have already assumed n\in Z as the first case to be considered. I am kind of tired at the moment to do the remaining maths.
    As long as n\in\mathbb{Z} you are fine. But the definition of \gamma isn't for some fixed n, so if you want to exploit it in any viable way you must let n\to\infty in which case x is no longer an integer.
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    (Original post by Lord of the Flies)
    As long as n\in\mathbb{Z} you are fine. But the definition of \gamma isn't for some fixed n, so if you want to exploit it in any viable way you must let n\to\infty in which case x is no longer an integer.
    How about of n\notin Z, then x\notin Z (Just a claim), can it do the same trick?

    EDIT: But isn't this summation only accept n\in N(I am not going to consider measure theory)?
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    (Original post by agostino981)
    How about of n\notin Z, then x\notin Z (Just a claim), can it do the same trick?
    No, because then your definition of \gamma would make absolutely no sense

    Out of interest, was this attempt inspired by a famous proof of the irrationality of e?
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    (Original post by Lord of the Flies)
    No, because then your definition of \gamma would make absolutely no sense

    Out of interest, was this attempt inspired by a famous proof of the irrationality of e?
    But isn't the summation in \gamma forces n to be natural number?

    Indeed. And I used the same trick to prove \sum_{0\le n \le \infty}\frac{(-1)^k}{(k+1)^{k+1}} to be irrational. No limit on that one so nothing happens but contradiction, so it is irrational.
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    (Original post by agostino981)
    But isn't the summation in \gamma forces n to be natural number?
    It is an infinite series over \mathbb{N}. I was referring to the sum not making sense if n were not an integer.
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    (Original post by Lord of the Flies)
    It is an infinite series over \mathbb{N}. I was referring to the sum not making sense if n were not an integer.
    So n is a natural number, so x should be an integer.

    EDIT: Measure theory probably works at this moment if n is not a natural number.
 
 
 
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