e is irrational Prove that e is an irrational number.
Recall that $\,\mathrm{e}=\displaystyle\sum_{n=0}^\infty\frac{1}{n!},\,\,$ and assume $\,\mathrm{e}\,$ is rational, then
$$\sum\limits_{k=0}^\infty \frac{1}{k!} = \frac{a}{b},\quad \text{for some positive integers}\,\,\, a,b.$$
so
$$b\sum\limits_{k=0}^\infty \frac{1}{k!} =a$$
or
$$
b\left(1+1+\frac{1}{2} + \frac{1}{6} +\cdots \right)= a.
$$
Where can I go from here?

Here is yet another one, which is one of my favorite irrationality/transcendence proofs :
The confluent hypergeometreic series 
$$_{0}F_{1}(k; z) = \sum_{n = 0}^\infty \frac1{(k)_n} \frac{z^n}{n!}$$
Satisfies the more-or-less easily verifiable identity 
$$_0F_1(k-1;z) - {}_0F_1(k; z) = \frac{z}{k(k-1)}{}_0F_1(k+1;z)$$
Iterating this, one ends up with the continued fraction 
$$\frac{{}_0F_1(k+1;z)}{k{}_0F_1(k;z)} = \frac1{k+\cfrac{z}{k+1+\cfrac{z}{k+2+\cfrac{z}{k+3+\cdots}}}}$$ 
Now note that $_0F_1(3/2;x^2/4) = \cosh(x)$ and $x \,{}_0F_1(3/2;x^2/4) = \sinh(x)$, hence applying the above one has the pretty well-known continued fraction
$$\tanh(x) = \cfrac{x/2}{\frac{1}{2} + \cfrac{\frac{x^2}{4}}{\frac{3}{2} + \cfrac{\frac{x^2}{4}}{\frac{5}{2} + \cfrac{\frac{x^2}{4}}{\frac{7}{2} + \cdots}}}} = \cfrac{x}{1 + \cfrac{x^2}{3 + \cfrac{x^2}{5 + \cfrac{x^2}{7 + \cdots}}}}$$
Note, however, that $$\tanh(x) = \frac{\exp(x) - \exp(-x)}{\exp(x) + \exp(-x)}$$ but since the continued fraction above is not finite, $\tanh(1)$ is not rational. Hence $e$ is not rational either.
