Page:Calculus Made Easy.pdf/161

It is, however, worth while to find another way of calculating this immensely important figure.

Accordingly, we will avail ourselves of the binomial theorem, and expand the expression $\left(1+{\dfrac {1}{n}}\right)^{n}$ in that well-known way.

The binomial theorem gives the rule that

${\begin{aligned}(a+b)^{n}&=a^{n}+n{\dfrac {a^{n-1}b}{1!}}+n(n-1){\dfrac {a^{n-2}b^{2}}{2!}}\\&{\phantom {=a^{n}\ }}+n(n-1)(n-2){\dfrac {a^{n-3}b^{3}}{3!}}+{\text{etc}}.\\\end{aligned}}$

Putting $a=1$ and $b={\dfrac {1}{n}}$, we get

${\begin{aligned}\left(1+{\dfrac {1}{n}}\right)^{n}&=1+1+{\dfrac {1}{2!}}\left({\dfrac {n-1}{n}}\right)+{\dfrac {1}{3!}}{\dfrac {(n-1)(n-2)}{n^{2}}}\\&{\phantom {=1+1\ }}+{\dfrac {1}{4!}}{\dfrac {(n-1)(n-2)(n-3)}{n^{3}}}+{\text{etc}}.\end{aligned}}$

Now, if we suppose n to become indefinitely great, say a billion, or a billion billions, then $n-1$, $n-2$, and $n-3$, etc., will all be sensibly equal to $n$; and then the series becomes

$\epsilon =1+1+{\dfrac {1}{2!}}+{\dfrac {1}{3!}}+{\dfrac {1}{4!}}+{\text{etc}}.\ldots$

By taking this rapidly convergent series to as many terms as we please, we can work out the sum to any desired point of accuracy. Here is the working for ten terms: