Determine convergence or divergence of the alternating series.
-$$\sum _ { n = 1 } ^ { \infty } \frac { ( - 2 ) ^ { n } } { 4 n ^ { 4 } + 8 ^ { n } }$$
A) Converges
B) Diverges

Question

Quizplus · Accepted Answer

Note that the series is alternating, so it suffices to show that the absolute value of the terms goes to 0. We have 
$$\left| \frac{(-2)^n}{4n^4 + 8^n} \right| = \frac{2^n}{4n^4 + 8^n} \leq \frac{2^n}{8^n} = \left( \frac{1}{4} \right)^n$$
Since the series $\sum_{n=1}^{\infty} \left( \frac{1}{4} \right)^n$ converges (it is a geometric series with common ratio $\frac{1}{4}$ and first term $\frac{1}{4}$), we have by the Comparison Test that our original series $\sum_{n=1}^{\infty} \frac{(-2)^n}{4n^4 + 8^n}$ converges.

Determine Convergence or Divergence of the Alternating Series ∑n=1∞(−2)n4n4+8n\sum _ { n = 1 } ^ { \infty } \frac { ( - 2 ) ^ { n } } { 4 n ^ { 4 } + 8 ^ { n } }n=1∑∞​4n4+8n(−2)n​

Determine Convergence or Divergence of the Alternating Series $\sum _ { n = 1 } ^ { \infty } \frac { ( - 2 ) ^ { n } } { 4 n ^ { 4 } + 8 ^ { n } }$