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The Euler-Mascheroni constant gamma is defined as:

$\gamma=\lim\limits_{n \rightarrow \infty}\left(\sum\limits_{m=1}^{n} \frac{1}{m} - \log(n)\right)$

From this previous question Do these series converge to logarithms? $\log(n)$ can be written:

$\log(n)=\sum\limits_{k=1}^\infty \sum\limits_{a=1}^{n-1}\frac{1}{kn-a}-\sum\limits_{k=1}^\infty\frac{n-1}{kn}$

$\gamma=\lim\limits_{n \rightarrow \infty}\left(\sum\limits_{m=1}^{n} \frac{1}{m} - \left(\sum\limits_{k=1}^\infty \sum\limits_{a=1}^{n-1}\frac{1}{kn-a}-\sum\limits_{k=1}^\infty\frac{n-1}{kn}\right)\right)$

$\gamma=\lim\limits_{n \rightarrow \infty}\left(\sum\limits_{m=1}^{n} \frac{1}{m} - \sum\limits_{k=1}^\infty \sum\limits_{a=1}^{n-1}\frac{1}{kn-a} + \sum\limits_{k=1}^\infty\frac{n-1}{kn}\right)$

Can this last expression be further simplified?

  • 0
    @GEdgar: That is $p$robably right.2011-06-21

3 Answers 3

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Careful, we don't actually have

$\log(n)=\sum\limits _{k=1}^{\infty}\sum\limits _{a=1}^{n-1}\frac{1}{kn-a}-\sum\limits _{k=1}^{\infty}\frac{n-1}{kn},$

since neither series converges. Instead we have

$\log(n)=\lim_{M\rightarrow\infty}\sum\limits _{k=1}^{M}\sum\limits _{a=1}^{n-1}\frac{1}{kn-a}-\sum\limits _{k=1}^{M}\frac{n-1}{kn}.$

Now, since

$\sum\limits _{k=1}^{M}\frac{n-1}{kn}=\sum\limits _{k=1}^{M}\frac{1}{k}-\sum\limits _{k=1}^{M}\frac{1}{kn}$

we can rewrite

$\log(n)=\lim_{M\rightarrow\infty}\sum\limits _{k=1}^{M}\sum\limits _{a=0}^{n-1}\frac{1}{kn-a}-\sum\limits _{k=1}^{M}\frac{1}{k}=\lim_{M\rightarrow\infty}\sum_{k=1}^{nM}\frac{1}{k}-\sum_{k=1}^{M}\frac{1}{k}.$

Thus we have

$\gamma=\lim_{n\rightarrow\infty}\lim_{M\rightarrow\infty}\left(\sum_{k=1}^{n}\frac{1}{k}-\sum_{k=1}^{nM}\frac{1}{k}+\sum_{k=1}^{M}\frac{1}{k}\right).$

Personally, I quite like this limit since it has a nice symmetry when we switch the order of the limits. Also, it generalizes to give limits over l variables which are invariant under permutation. Let $H_{k}$ be the $k^{th}$ harmonic number. Then the above was

$\gamma=\lim_{n\rightarrow\infty}\lim_{M\rightarrow\infty}\left(H_{n}-H_{nm}+H_{m}\right).$

Here is $l=3$:

$\gamma=\lim_{n_{1}\rightarrow\infty}\lim_{n_{2}\rightarrow\infty}\lim_{n_{3}\rightarrow\infty}\left(\left(H_{n_{1}}+H_{n_{2}}+H_{n_{3}}\right)-\left(H_{n_{1}n_{2}}+H_{n_{2}n_{3}}+H_{n_{3}n_{1}}\right)+\left(H_{n_{1}n_{2}n_{3}}\right)\right).$

In general

$\gamma=\lim_{n_{1}\rightarrow\infty}\cdots\lim_{n_{l}\rightarrow\infty}\left(\sum_{i_{1}=1}^{l}H_{n_{i_{1}}}-\sum_{i_{1}

So I guess it depends on what you mean by “simplify further.” I find the form

$\gamma=\lim_{n\rightarrow\infty}\lim_{m\rightarrow\infty}\left(H_{n}-H_{nm}+H_{m}\right)$

to be quite simple.

Hope that helps,

  • 0
    Yeah, it does, because that's equal to:$\lim_{n_1\to\infty}\lim_{n_2\to\infty}\lim_{n_3\to\infty}\\ \frac12\big((H_{n_1}-\ln n_1)+(H_{n_2}-\ln n_2)+(H_{n_3}-\ln n_3)-(H_{n_1n_2n_3}-\ln n_1n_2n_3)\big)$which goes to $\frac12(\gamma+\gamma+\gamma-\gamma)=\gamma$. Same strategy proves the similar limits in the answer.2016-01-26
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@Eric can you give the proof that this is equal (for every possible value of n): $\ln(n)=\lim_{M\rightarrow\infty}\sum\limits _{k=1}^{M}\sum\limits _{a=1}^{n-1}\frac{1}{kn-a}-\sum\limits _{k=1}^{M}\frac{n-1}{kn}.$

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Setting $m=n$ in Naslund's equation we get

$ \gamma= \sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right) $

This may be used to write a closed form for the digamma function, see Re-Expressing the Digamma.