# How do you calculate Euler's Number?

Jul 30, 2017

Note
This solution provides an estimate for Euler's Number (or $e$, the base of Natural (or Naperian) logarithms) rather than Euler's Constant.

There may be differing opinions, but I would of thought that simplest derivation is to use the Maclaurin Series for ${e}^{x}$

The Maclaurin Series for ${e}^{x}$ is as follows:

 e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + (x^5)/(5!) + ...

So the power series sought for $e$ can be gained by putting $x = 1$, viz:

 e = 1 + 1 + (1)/(2!) + (1)/(3!) + (1)/(4!) + (1)/(5!) + ...

We can compute an approximating fraction by truncating the series as we see fit. For example, If we restrict ourself to the first five terms, we get:

 e ~~ 1 + 1 + (1)/(2!) + (1)/(3!) + (1)/(4!)
$\setminus \setminus \setminus = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24}$
$\setminus \setminus \setminus = \frac{65}{24}$
$\setminus \setminus \setminus = 2.708333$

We can compare this to a calculator answer for the Euler Constant, $e$:

$e = 2.7182818 \ldots$

And for further comparison, if we take further terms we get:

$6$ terms: $e \approx \frac{65}{24} + \frac{1}{120}$
$\text{ } = \frac{163}{60}$
$\text{ } = 2.716666 \ldots$

$7$ terms: $e \approx \frac{163}{60} + \frac{1}{720}$
$\text{ } = \frac{1957}{720}$
$\text{ } = 2.718055 \ldots$

$8$ terms: $e \approx \frac{1957}{720} + \frac{1}{5040}$
$\text{ } = \frac{685}{252}$
$\text{ } = 2.718253 \ldots$

Jul 30, 2017

The Euler-Mascheroni constant is defined as:

$\gamma = {\lim}_{n \to \infty} \left({\sum}_{k = 1}^{n} \frac{1}{k} - \ln n\right)$

that is as the difference between the $n$-th partial sum of the harmonic series and $\ln n$.

Since the limit is in indeterminate form we must prove it is convergent. Write $\ln n$ as:

$\ln n = \ln n - \ln \left(n - 1\right) + \ln \left(n - 1\right) - \ldots - \ln 1 = {\sum}_{k = 1}^{n - 1} \left(\ln \left(k + 1\right) - \ln k\right) = {\sum}_{k = 1}^{n - 1} \ln \left(1 + \frac{1}{k}\right)$

Then:

$\gamma = {\sum}_{k = 1}^{\infty} \frac{1}{k} - \ln \left(1 + \frac{1}{k}\right)$

Now note that using Taylor's formula with Lagrange's rest and $0 < x < 1$:

$\ln \left(1 + x\right) = {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {x}^{k + 1} / \left(k + 1\right) = x - {\xi}^{2} / 2$ with $\xi \in \left(0 , x\right)$

So:

$\frac{1}{k} - \ln \left(1 + \frac{1}{k}\right) = {\xi}^{2} / 2$ with $\xi \in \left(0 , \frac{1}{k}\right)$

and:

$0 < \frac{1}{k} - \ln \left(1 + \frac{1}{k}\right) < \frac{1}{2 {k}^{2}}$

As the series:

${\sum}_{k = 1}^{\infty} \frac{1}{2 {k}^{2}}$

is convergent based on the $p$-series test, then by direct comparison also the series:

$\gamma = {\sum}_{k = 1}^{\infty} \frac{1}{k} - \ln \left(1 + \frac{1}{k}\right)$

is convergent and $\gamma$ is a finite real number.

Incidentally this proves that the partial sums of the harmonic series diverge with the same order as $\ln n$