Euler-Mascheroni constant expression, further simplification

Question

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?

Answer 1

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,

Answer 2

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

Answer 3

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.