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

Jan 14, 2017

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 $\sum {a}_{n}$, we can make a determination about its convergence by taking $L = {\lim}_{a \rightarrow \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid$. Examine the value of $L$:

• If $L > 1$, then $\sum {a}_{n}$ is divergent.
• If $L = 1$, then the test is inconclusive.
• If $L < 1$, then $\sum {a}_{n}$ 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.