# Thoughts on Infinity (Part 3b)

The five most important numbers in mathematics are $0$, $1$, $e$, $\pi$, and $i$. In sixth place (a distant sixth place) is probably $\gamma$, the Euler-Mascheroni constant. See Mathworld or Wikipedia for more details. (For example, it’s astounding that we still don’t know if $\gamma$ is irrational or not.)

In yesterday’s post, we’ve seen the curious phenomenon that the commutative and associative laws do not apply to a conditionally convergent series or infinite product. In tomorrow’s post, I’ll present another classic example of this phenomenon due to Cauchy. However, to be ready for this fact, I’ll need to see how $\gamma$ arises from a certain conditionally convergent series.

Separately define the even and odd terms of the sequence $\{a_n\}$ by

$a_{2n} = \displaystyle \int_n^{n+1} \frac{dx}{x}$

and

$a_{2n-1} = \displaystyle \frac{1}{n}$.

It’s pretty straightforward to show that this sequence is decreasing. The function $f(x) = \displaystyle \frac{1}{x}$ is clearly decreasing for $x > 0$, and so the maximum value of $f(x)$ on the interval $[n,n+1]$ must occur at the left endpoint, while the minimum value must occur at the right endpoint. Since the length of this interval is $1$, we have

$\displaystyle \frac{1}{n+1} \cdot 1 < \displaystyle \int_n^{n+1} \frac{dx}{x} < \displaystyle \frac{1}{n} \cdot 1$,

or

$a_{2n+1} < a_{2n} < a_{2n-1}$.

Since the subsequence $\{a_{2n-1}\}$ clearly decreases to $0$, this shows the full sequence $\{a_n\}$ is a decreasing sequence with limit $0$.

By the alternating series test, this implies that the series

$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} a_n$

converges. This limit is called the

Since this series converges, that means that the limit of the partial sums converges to $\gamma$:

$\displaystyle \lim_{M \to \infty} \sum_{n=1}^M (-1)^{n-1} a_n = \gamma$.

Let’s take the upper limit to be an odd number $M$, where $M = 2N-1$ and $N$ is an integer. Then by separating the even and odd terms, we obtain

$\displaystyle \sum_{n=1}^{2N-1} (-1)^{n-1} a_n = \displaystyle \sum_{n=1}^{N} (-1)^{2n-1-1} a_{2n-1} + \sum_{n=1}^{N-1} (-1)^{2n-1} a_{2n}$

$= \displaystyle \sum_{n=1}^N a_{2n-1} - \sum_{n=1}^{N-1} a_{2n}$

$= \displaystyle \sum_{n=1}^N \frac{1}{n} - \sum_{n=1}^{N-1} \int_n^{n+1} \frac{dx}{x}$

$= \displaystyle \sum_{n=1}^N \frac{1}{n} - \int_1^N \frac{dx}{x}$.

Therefore,

$\displaystyle \lim_{N \to \infty} \left( \sum_{n=1}^N \frac{1}{n} - \int_1^N \frac{dx}{x} \right) = \gamma$.

With this interpretation, the sum can be viewed as the sum of the $N$ rectangles in the above picture, while the integral is the area under the hyperbola. Therefore, the limit $\gamma$ can be viewed as the limit of the blue part of the above picture.

In other words, it’s an amazing fact that while both

$\displaystyle \sum_{n=1}^\infty \frac{1}{n}$

and

$\displaystyle \int_1^\infty \frac{dx}{x}$

diverge, somehow the difference

$\displaystyle \lim_{N \to \infty} \left(\sum_{n=1}^N \frac{1}{n} - \int_1^N \frac{dx}{x} \right)$

converges… and this limit is defined to be the number $\gamma$.

## 24 thoughts on “Thoughts on Infinity (Part 3b)”

1. The difference function 1/x – 1/(x+1) encloses the blue bits, and its integral is INT(1/(x(x+1)_, which is less than INT(1/(x^2), which inturn is finite when taken from 0 to infinity. Hence convergence of the sum of blue bits.