Determine convergence or divergence of the series.
-$$\sum _ { n = 1 } ^ { \infty } n ^ { 2 } e ^ { - 7 n ^ { 8 } }$$
A) Diverges
B) Converges

Question

Quizplus · Accepted Answer

To test for convergence, we can use the ratio test:
$$\lim_{n\rightarrow\infty} \left|\frac{(n+1)^2 e^{-7(n+1)^8}}{n^2 e^{-7n^8}}\right| = \lim_{n\rightarrow\infty} \left|\frac{(n+1)^2}{n^2} e^{-7(n^8+8n^7+28n^6+\cdots)}\right| = \lim_{n\rightarrow\infty} \left|\frac{(n+1)^2}{n^2} e^{-\infty}\right| = \lim_{n\rightarrow\infty} \left|\frac{(n+1)^2}{n^2}\right| = 1$$
Since the limit is equal to 1, the series does not converge by the ratio test. Therefore, the series diverges.

Determine Convergence or Divergence of the Series ∑n=1∞n2e−7n8\sum _ { n = 1 } ^ { \infty } n ^ { 2 } e ^ { - 7 n ^ { 8 } }n=1∑∞​n2e−7n8

Determine Convergence or Divergence of the Series $\sum _ { n = 1 } ^ { \infty } n ^ { 2 } e ^ { - 7 n ^ { 8 } }$