Proof of Euler-Mascheroni Constant

1. Oct 19, 2011

coki2000

Hi PF,

I have question about harmonic series.

$$\lim_{n\to \infty}(\sum_{k=1}^{\infty}\frac{1}{k}-ln(n))=ln(n)+\gamma +\epsilon _0$$

I couldn't find any proof for that equality. Do you know its exact proof?

Thanks for helps.

2. Oct 19, 2011

dextercioby

The formula is wrong. It should read sum to n in the brackets and no ln(n) in the RHS.

3. Oct 19, 2011

coki2000

Oh, sorry you're right it should be like that

$$\sum_{k=1}^{n}\frac{1}{k}=ln(n)+\gamma +\epsilon _n$$

My question is where it comes from? How can we prove it?

4. Oct 19, 2011

Citan Uzuki

Essentially, it comes from the fact that γ is defined to be $\lim_{n \rightarrow \infty}\sum_{k=1}^n \frac{1}{k} - \ln(n)$, so the only thing that is necessary is to prove that this limit actually exists and is finite. For this, consider the following:

\begin{align*} \sum_{k=1}^{n} \frac{1}{k} - \ln(n) &= \sum_{k=1}^{n} \int_{k}^{k+1}\frac{1}{\lfloor x \rfloor} \ dx - \int_{1}^{n} \frac{1}{x} \\ &= \int_{1}^{n}\frac{1}{\lfloor x \rfloor} - \frac{1}{x} \ dx + \int_{n}^{n+1} \frac{1}{\lfloor x \rfloor} \ dx \\ &= \int_{1}^{n} \frac{x-\lfloor x \rfloor}{x \lfloor x \rfloor} \ dx + \frac{1}{n} \end{align*}

Now, as n→∞, 1/n → 0, and the integral on the left converges to:

$$\int_{1}^{\infty} \frac{x-\lfloor x \rfloor}{x \lfloor x \rfloor} \ dx$$

But this integral is finite, since:

$$\frac{x-\lfloor x \rfloor}{x \lfloor x \rfloor} \leq \frac{1}{\lfloor x \rfloor^{2}}$$

And:

$$\int_{1}^{\infty} \frac{1}{\lfloor x \rfloor^{2}} \ dx = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} < \infty$$

5. Oct 19, 2011

coki2000

@Citan Uzuki

Thank you very much for your enlightening response. But what about the term $\epsilon_n$. Where it comes from? And also how can we calculate γ numerically?

6. Oct 19, 2011

dextercioby

The epsilon is a very small number accounting for the error which disappears when taking the limit.

7. Oct 19, 2011

coki2000

In this http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence", it says that $\epsilon_k$ is approximately 1/2k so I wonder why Euler inserted this factor into this equality and how he found that it is approximately 1/2k(not 1/5k for example). And is there any special name for that epsilon factor?

Thanks for all your helps :)

Last edited by a moderator: Apr 26, 2017