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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?

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Perhaps it is not good to have a divergent series like $\sum\limits_{k=1}^\infty\frac{n-1}{kn}$ inside your formula. – GEdgar Jun 21 '11 at 15:17
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. – GEdgar Jun 21 '11 at 15:32
@GEdgar: That is probably right. – Mats Granvik Jun 21 '11 at 16:28

2 Answers 2

up vote 8 down vote accepted

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


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


Here is $l=3$:


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


to be quite simple.

Hope that helps,

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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". – Mats Granvik Jun 21 '11 at 16:20

@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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