Question

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

Answer 1

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 rarr infty)abs(a_(n+1)/a_n)=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 rarr infty)abs((-2)/(n+1))`
`0`

`0<1` so the series converges absolutely.