# How do you test the alternating series Sigma ((-1)^(n+1)2^n)/(n!) from n is [0,oo) for convergence?

Apr 24, 2018

Converges absolutely

#### Explanation:

Use the ratio test. This test works for alternating series.

If there is a series ${\sum}_{n = 0}^{\infty} {a}_{n}$ then consider ${\lim}_{n \rightarrow \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = L$.
If $L < 1$ then the series absolutely converges.
If $L > 1$ then the series diverges.
If $L = 1$ then the test is inconclusive.

For our series, we have:
lim_(n rarr infty)abs(((-1)^((n+1)+1)2^(n+1))/((n+1)!)div((-1)^(n+1)2^n)/(n!))
lim_(n rarr infty)abs(((-1)^(n+2)2^(n+1)n!)/((-1)^(n+1)2^n(n+1)!)
${\lim}_{n \rightarrow \infty} \left\mid \frac{- 2}{n + 1} \right\mid$
$0$

$0 < 1$ so the series converges absolutely.