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Check all of the following that are true for the series ∑n=1∞(−2)n(n!) A. This series converges B. This series diverges C. The integral test can be used to determine convergence of this series. D. The comparison test can be used to determine convergence of this series. E. The limit comparison test can be used to determine convergence of this series. F. The ratio test can be used to determine convergence of this series. G. The alternating series test can be used to determine convergence of this series. Note: You can earn partial credit on this prob

Answer :

LammettHash

I assume the series is

[tex]\displaystyle\sum_{n=1}^\infty\frac{(-2)^n}{n!}[/tex]

By the ratio test, the series converges because

[tex]\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-2)^{n+1}}{(n+1)!}}{\frac{(-2)^n}{n!}}\right|=2\lim_{n\to\infty}\frac1{n+1}=0[/tex]

so A and F are true.

The integral test cannot be used because the summand is not monotonic (decreasing).

The comparison tests cannot be used because the series is alternating; these tests require the summand be non-negative.

The alternating series test can be used because

[tex]\dfrac{(-2)^n}{n!}=(-1)^n\dfrac{2^n}{n!}[/tex]

and [tex]\dfrac{2^n}{n!}\ge0[/tex]. For large enough [tex]n[/tex], we have [tex]2^n<n![/tex], and [tex]\dfrac{2^n}{n!}\to0[/tex]. Then by the alternating series test, the series converges. So G is also true.

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