## Existence of Euler–Mascheroni constant

One of the things that baffled me when I was learning calculus was that harmonic series

$\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$

diverges. Clearly, the terms are getting smaller and smaller (and in fact the terms go to zero in the limit the number of terms is taken to infinity). Yet, the series diverges. There is more than one way of proving this fact, and one of them is to find a function that approximates the growth rate of this series. It turns out that $\log x$ works for this purpose (where $\log$ denotes natural logarithm). More concretely,

Proposition. There exists a constant $\gamma$ such that

$\gamma=\displaystyle \lim\limits_{N\to\infty}\left(\sum\limits_{n=1}^{N}\frac{1}{n}-\log N\right)$

The constant $\gamma$ is known as Euler–Mascheroni constant. Approximately, $\gamma\approx 0.57721$.

Proof: We have

$\displaystyle \sum\limits_{n=1}^{N}\frac{1}{n}-\log N=\sum\limits_{n=1}^{N}\frac{1}{n}-\int\limits_{1}^{N}\frac{1}{x}\ \mathrm{d}x=\sum\limits_{n=1}^{N-1}\int\limits_{n}^{n+1}\left(\frac{1}{n}-\frac{1}{x}\right)\mathrm{d}x+\frac{1}{N}$

Observe that for all $x$ satisfying $n\le x \le n+1$, the following inequality holds:

$\displaystyle 0\le \frac{1}{n}-\frac{1}{x}\le \frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}<\frac{1}{n^2}$

so that

$\displaystyle \left|\frac{1}{n}-\frac{1}{x}\right|<\frac{1}{n^2}$

for all $n\le x \le n+1$. But then,

$\displaystyle \sum\limits_{n=1}^{N}\frac{1}{n}-\log N=\sum\limits_{n=1}^{N}a_n + \frac{1}{N}$

where $|a_n|<1/n^2$. Since the sum $\sum \frac{1}{n^2}$ converges and $\frac{1}{N}\to 0$ as $N\to\infty$, it follows that the left hand side of the above equation converges, and that the supposed limit $\gamma$ exists. $\blacksquare$