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Use the Ratio Test to Investigate the Series n=12nn!\sum_{n=1}^{\infty} \frac{2^{n}}{n !}

Question 21

Multiple Choice

Use the ratio test to investigate the series n=12nn!\sum_{n=1}^{\infty} \frac{2^{n}}{n !} .


A) limn2n+1(n+1) !2nn!=limn2n+1=0\lim _{n \rightarrow \infty} \frac{\frac{2^{n+1}}{(n+1) !}}{\frac{2 n}{n !}}=\lim _{n \rightarrow \infty} \frac{2}{n+1}=0 , so the series n=12nn!\sum_{n=1}^{\infty} \frac{2^{n}}{n !} diverges.
B) limn2n+1(n+1) !2nn!=limn2n+1=0<1\lim _{n \rightarrow \infty} \frac{\frac{2^{n+1}}{(n+1) !}}{\frac{2 n}{n !}}=\lim _{n \rightarrow} \frac{2}{n+1}=0<1 , so the series n=12nn!\sum_{n=1}^{\infty} \frac{2^{n}}{n !} converges.
C) limn2n+1(n+1) !2nn!=limn2n+1=0\lim _{n \rightarrow \infty} \frac{\frac{2^{n+1}}{(n+1) !}}{\frac{2 n}{n !}}=\lim _{n \rightarrow \infty} \frac{2}{n+1}=0 , so the ratio test fails to tell us anything about the series.
D) limn2nn!2n+1(n+1) !=limnn+12=\lim _{n \rightarrow \infty} \frac{\frac{2 n}{n !}}{\frac{2^{n+1}}{(n+1) !}}=\lim _{n \rightarrow \infty} \frac{n+1}{2}=\infty , so the series n=12nn!\sum_{n=1}^{\infty} \frac{2^{n}}{n !} converges.

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