Absolute Convergence and Divergence — Question 7

*Question 7 Consider $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{n!}$.

1. Apply the Ratio Test to the absolute-value series.

2. Classify convergence.

3. Identify the exact sum using the exponential series.

*Question 7 – Solution

Let $b_n=2^n/n!$. Then $$\frac{b_{n+1}}{b_n}=\frac{2}{n+1}\longrightarrow0<1.$$ The Ratio Test proves that $\sum b_n$ converges.

Therefore the signed series converges absolutely. More generally, replacing $2$ by any fixed real base $c$ gives ratio $|c|/(n+1)\to0$.

From $e^x=\sum_{n=0}^{\infty}x^n/n!$, substitute $x=-2$: $$\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{n!}=e^{-2}.$$