Question

use any method to determine if the series converges or diverges. give reasons for your answer.

sum_{n = 1}^{infty} (-1)^n n^2 e^{-12n}

select the correct choice below and fill in the answer box to complete your choice. (type an exact answer.)

a. the series diverges because the limit found using the nth-term test is

b. the series diverges because the limit found using the ratio test is

c. the series converges because the limit found using the ratio test is

d. the series converges because the limit found using the nth-term test is

Explanation:

Step1: Apply the nth - Term Test

The nth - Term Test states that for a series \(\sum_{n = 1}^{\infty}a_{n}\), if \(\lim_{n
ightarrow\infty}a_{n}
eq0\), then the series \(\sum_{n = 1}^{\infty}a_{n}\) diverges.
For the series \(\sum_{n = 1}^{\infty}(- 1)^{n}(n^{2}-12n)e^{n}\), we find \(\lim_{n
ightarrow\infty}\vert a_{n}\vert=\lim_{n
ightarrow\infty}(n^{2}-12n)e^{n}\).
We use L'Hopital's Rule. Let \(y=(n^{2}-12n)e^{n}\). We can also consider the fact that the exponential function \(e^{n}\) grows faster than any polynomial function.
As \(n
ightarrow\infty\), the polynomial function \(n^{2}-12n\) is positive for large \(n\) (since \(n^{2}-12n=n(n - 12)\) and for \(n>12\), \(n^{2}-12n>0\)).
We know that \(\lim_{n
ightarrow\infty}n^{k}e^{n}=\infty\) for \(k>0\). Here \(k = 2\) (from \(n^{2}\) in \(n^{2}-12n\)). So \(\lim_{n
ightarrow\infty}(n^{2}-12n)e^{n}=\infty\).
Since \(\lim_{n
ightarrow\infty}(-1)^{n}(n^{2}-12n)e^{n}\) does not exist (because \(\lim_{n
ightarrow\infty}\vert(-1)^{n}(n^{2}-12n)e^{n}\vert=\infty\)), and in particular \(\lim_{n
ightarrow\infty}(-1)^{n}(n^{2}-12n)e^{n}
eq0\).

Answer:

A. The series diverges because the limit found using the nth - Term Test is \(\infty\) (i.e., non - zero).