4
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ Perhaps it is not good to have a divergent series like $\sum\limits_{k=1}^\infty\frac{n-1}{kn}$ inside your formula. $\endgroup$
    – GEdgar
    Jun 21, 2011 at 15:17
  • $\begingroup$ That is, you want $$\log(n)=\sum\limits_{k=1}^\infty \left[\sum\limits_{a=1}^{n-1}\frac{1}{kn-a}-\frac{n-1}{kn}\right]$$ not two separate sums. $\endgroup$
    – GEdgar
    Jun 21, 2011 at 15:32
  • $\begingroup$ @GEdgar: That is probably right. $\endgroup$ Jun 21, 2011 at 16:28

3 Answers 3

11
$\begingroup$

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

$$\gamma=\lim_{n_{1}\rightarrow\infty}\cdots\lim_{n_{l}\rightarrow\infty}\left(\sum_{i_{1}=1}^{l}H_{n_{i_{1}}}-\sum_{i_{1}<i_{2}\leq l}H_{n_{i_{1}}n_{i_{2}}}+\cdots+(-1)^{l}H_{n_{1}\cdots n_{l}}\right).$$

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,

$\endgroup$
3
  • $\begingroup$ Wow thanks, that is a really nice looking expression in the middle of your answer after "Thus we have". If I understood it right, the use of a double limit is the way to make it computable. My choice of words was not so good. I should have left out the word "further". $\endgroup$ Jun 21, 2011 at 16:20
  • $\begingroup$ Does:$$\lim_{n_1\to\infty}\lim_{n_2\to\infty}\lim_{n_3\to\infty}\frac12(H_{n_1}+H_{n_2}+H_{n_3}-H_{n_1n_2n_3})$$also equal $\gamma$? $\endgroup$ Jan 26, 2016 at 20:21
  • $\begingroup$ 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. $\endgroup$ Jan 26, 2016 at 20:29
1
$\begingroup$

@Eric can you give the proof that this is equal (for every possible value of n): $$\ln(n)=\lim_{M\rightarrow\infty}\sum\limits _{k=1}^{M}\sum\limits _{a=1}^{n-1}\frac{1}{kn-a}-\sum\limits _{k=1}^{M}\frac{n-1}{kn}.$$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .