Euler Constant

Prove that the sequence $\displaystyle{\sum_{i=1}^{n}}$ $\frac{1}{n}$ – log n is convergert . [And the constant is denoted by $\gamma$and it is still unknown wheather it is irrational or not ].

Solution :Monotony: It is easy to prove that the sequence is decreasing. Indeed:

$$\gamma_{n+1} - \gamma_n = \left ( \mathcal{H}_{n+1}- \ln (n+1) \right ) - \left ( \mathcal{H}_n - \ln n \right ) = \frac{1}{n}+ \ln \left ( 1- \frac{1}{n} \right )<0 $$
The inequality is valid since $\ln (1-x)$ is a concave function hence lies beneath the line $y=-x$ that is its tangent to its graph at $x_0=0$. Plugging $x=1/n$ yields the result.

Boundness: Using the fact that:

$$\mathcal{H}_n = \sum_{k=1}^{n}\frac{1}{k}> \int_{1}^{n}\frac{{\rm d}t}{t}= \ln (n+1)> \ln n$$
we have that $\gamma_n = \mathcal{H}_n -\ln n >0$. Hence $\gamma_n$ is bounded from below.

Therefore the sequence converges.

The limit of this sequence is a very well known constant, namely Euler – Mascheroni constant.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s