Question

How do you calculate Euler's Number?

Answer 1

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!)`
`\ \ \ = 1 + 1 + 1/2+1/6+1/24`
`\ \ \ = 65/24`
`\ \ \ = 2.708333`

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

`e = 2.7182818 ...`

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

`6` terms: `e~~65/24+1/120`
`" " = 163/60`
`" " = 2.716666 ...`

`7` terms: `e~~ 163/60 + 1/720`
`" " = 1957/720`
`" " = 2.718055 ...`

`8` terms: `e~~ 1957/720 + 1/5040`
`" " = 685/252`
`" " = 2.718253 ...`

Answer 2

The Euler-Mascheroni constant is defined as:

`gamma = lim_(n->oo) (sum_(k=1)^n 1/k -lnn)`

that is as the difference between the `n`-th partial sum of the harmonic series and `lnn`.

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

`lnn = ln n - ln(n-1) + ln(n-1) -...-ln1 = sum_(k=1)^(n-1) (ln(k+1) -lnk )= sum_(k=1)^(n-1) ln (1+1/k)`

Then:

`gamma = sum_(k=1)^oo 1/k-ln(1+1/k)`

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

`ln(1+x) = sum_(k=0)^oo (-1)^k x^(k+1)/(k+1) = x - xi^2/2` with `xi in (0,x)`

So:

`1/k-ln(1+1/k) = xi^2/2` with `xi in (0,1/k)`

and:

`0 < 1/k-ln(1+1/k) < 1/(2k^2)`

As the series:

`sum_(k=1)^oo 1/(2k^2)`

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

`gamma = sum_(k=1)^oo 1/k-ln(1+1/k)`

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 `lnn`