$\text {The series } \sum _ { n = 1 } ^ { \infty } \frac { 1 } { ( 6 n + 7 ) ^ { 3 } } \text { converges. }$

Question

Quizplus · Accepted Answer

We can use the comparison test to show that the given series converges. Note that for all $n\geq 1$, we have
$$
\frac{1}{(6n+7)^3}\leq \frac{1}{(6n)^3}=\frac{1}{216n^3}.
$$
Since the series $\sum_{n=1}^{\infty}\frac{1}{n^3}$ converges (this is a well-known result), it follows that $\sum_{n=1}^{\infty}\frac{1}{(6n+7)^3}$ converges by the comparison test.

The series ∑n=1∞1(6n+7)3 converges. \text {The series } \sum _ { n = 1 } ^ { \infty } \frac { 1 } { ( 6 n + 7 ) ^ { 3 } } \text { converges. }The series ∑n=1∞​(6n+7)31​ converges.

$\text {The series } \sum _ { n = 1 } ^ { \infty } \frac { 1 } { ( 6 n + 7 ) ^ { 3 } } \text { converges. }$