# Does a_n=1/(n!)  converge?

Dec 13, 2015

Yes.

lim_(n->oo)1/(n!) = 0

#### Explanation:

As a way of building intuition about a sequence of the form $f \frac{n}{g} \left(n\right)$

1. Consider if the rate of growth of one of the functions is much faster than that of the other. As a handy mental reference for growth rates of common functions:
n^n > n! > c^n > n^c > ln(n) > c
(where $c$ is a constant)
If one of the functions "dominates" the other, then that is all that matters for the end behavior.

2. Consider the end behavior looking at the dominating function.
If we only look at what happens to the dominating function as $n \to \infty$, then if the result is of the form (+-oo)/? or ?/0 then the sequence will diverge, and if the result is of the form 0/? or ?/(+-oo) then the sequence will converge to 0.

3. If neither function obviously dominates the other, see if an equivalent sequence is clearer. For example, although ${n}^{2}$ and ${n}^{3}$ belong to the same "group" above,
${n}^{2} / {n}^{3} = \frac{1}{{n}^{3} / {n}^{2}} = \frac{1}{n} \to \frac{1}{\infty} = 0$

If none of the above gives an obvious answer, then more work may be needed, often through trickier algebraic manipulation, or techniques such as the squeeze theorem or L'hospital's rule.

While the above is not always enough of a justification for deciding a limit, especially more complicated ones, there is little else required for the given sequence, as it is valid to say that if ${\lim}_{n \to \infty} f \left(n\right) = \pm \infty$ then ${\lim}_{n \to \infty} \frac{c}{f} \left(n\right) = 0$

Thus, as lim_(n->oo)n! = oo we have lim_(n->oo)1/(n!) = 0