Question

How do you test the series `Sigma 1/(n!)` from n is `[0,oo)` for convergence?

Answer 1

Use the ratio test to show the series' convergence.

Explanation:

We will use the ratio test. The ratio test says that the for the series `suma_n`, we can make a determination about its convergence by taking `L=lim_(ararroo)abs(a_(n+1)/a_n)`. Examine the value of `L`:

• If `L>1`, then `suma_n` is divergent.
• If `L=1`, then the test is inconclusive.
• If `L<1`, then `suma_n` is (absolutely) convergent.

So for the series `sum_(n=0)^oo1/(n!)` we let `a_n=1/(n!)`. Then we see that

`L=lim_(nrarroo)abs((1/((n+1)!))/(1/(n!)))=lim_(nrarroo)abs((n!)/((n+1)!))`

This takes recalling a little bit about factorial. The definition of factorial states that `(n+1)! =(n+1)(n!)`, similar to how `7! = 7*6!`. Thus:

`L=lim_(nrarroo)abs((n!)/((n+1)(n!)))=lim_(nrarroo)abs(1/(n+1))=0`

Since `L=0` and therefore `L<1`, we see that `suma_n=sum_(n=0)^oo1/(n!)` is convergent through the ratio test.