Use the Limit Comparison Test to determine the convergence or divergence of the series $\sum _ { n = 6 } ^ { \infty } \frac { 1 } { n ^ { 7 } - 6 }$.
A)
The series $\sum _ { n = 6 } ^ { \infty } \frac { 1 } { n ^ { 2 } - 6 }$ diverges.
B)
The series $\sum _ { n = 6 } ^ { \infty } \frac { 1 } { n ^ { 7 } - 6 }$ converges.

Question

Quizplus · Accepted Answer

Using the Limit Comparison Test, we compare the given series to a known series $\sum \frac{1}{n^7}$:

\begin{align*}
\lim_{n	o\infty}\frac{\frac{1}{n^7-6}}{\frac{1}{n^7}} &= \lim_{n	o\infty}\frac{n^7}{n^7-6}\
&= \lim_{n	o\infty}\frac{1}{1-\frac{6}{n^7}}\
&= 1
\end{align*}

Since the limit exists and is finite, and the comparing series $\sum\frac{1}{n^7}$ converges (by the $p$-series test with $p=7>1$), we can conclude that the original series also converges by the Limit Comparison Test. Therefore, the answer is (B).

Use the Limit Comparison Test to Determine the Convergence or Divergence