Determine convergence or divergence of the series.
-$$\sum _ { n = 1 } ^ { \infty } \frac { \ln ( 2 n ) } { n ^ { 5 } }$$
A) Converges
B) Diverges

Question

Quizplus · Accepted Answer

We can use the limit comparison test with the series $\sum_{n=1}^\infty \frac{1}{n^5}$, which we know to converge by the $p$-series test. Taking the limit as $n\to\infty$ of the ratio $\frac{\ln(2n)}{n^5}\cdot\frac{n^5}{1}=\ln(2n)$, we get:
$$\lim_{n\to\infty}\ln(2n)=\infty$$
Since $\ln(2n)$ approaches infinity, which is a positive constant, as $n$ approaches infinity, the limit comparison test tells us that $\sum_{n=1}^\infty\frac{\ln(2n)}{n^5}$ and $\sum_{n=1}^\infty\frac{1}{n^5}$ either both converge or both diverge. Since we know that $\sum_{n=1}^\infty\frac{1}{n^5}$ converges, we conclude that $\sum_{n=1}^\infty\frac{\ln(2n)}{n^5}$ converges as well.

Determine Convergence or Divergence of the Series ∑n=1∞ln⁡(2n)n5\sum _ { n = 1 } ^ { \infty } \frac { \ln ( 2 n ) } { n ^ { 5 } }n=1∑∞​n5ln(2n)​

Determine Convergence or Divergence of the Series $\sum _ { n = 1 } ^ { \infty } \frac { \ln ( 2 n ) } { n ^ { 5 } }$