Find the values of p for which the series is convergent. $\sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { \left( \ln \left( n ^ { 6 } \right) \right) ^ { \gamma } }$ A) $p 1$ B) $p 0$ C) $p

Question

Find the values of p for which the series is convergent. $\sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { \left( \ln \left( n ^ { 6 } \right) \right) ^ { \gamma } }$ A) $p > 1$ B) $p > 0$ C) $p < 0$ D) $p < 1$

Quizplus · Accepted Answer

We can use the alternating series test to show that the series converges for all values of $\gamma>0$. Note that $\frac{1}{(\ln(n^6))^\gamma} = \frac{1}{(\gamma \ln(n))^{\gamma}} = \frac{1}{n^\gamma \ln^\gamma(n)}$. Let $a_n = \frac{1}{n^\gamma \ln^\gamma(n)}$. Then, $\lim_{n 	o \infty} a_n = 0$, and $a_{n+1} \leq a_n$ for all $n \geq 2$ (can be proven by taking derivatives). Thus, by the alternating series test, the series converges. Therefore, the answer is B.

Find the Values of P for Which the Series Is ∑n=2∞(−1)n(ln⁡(n6))γ\sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { \left( \ln \left( n ^ { 6 } \right) \right) ^ { \gamma } }∑n=2∞​(ln(n6))γ(−1)n​

Find the Values of P for Which the Series Is $\sum _ { n = 2 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { \left( \ln \left( n ^ { 6 } \right) \right) ^ { \gamma } }$