Determine whether the series is convergent or divergent. $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n ^ { 2 } - 6 n + 14 }$
A) converges
B) diverges

Question

Quizplus · Accepted Answer

We can use the comparison test with the p-series $\sum_{n=1}^{\infty}\frac{1}{n^2}$, which we know to converge. 
First, we complete the square in the denominator:
$n^2-6n+14 = (n-3)^2+5$
Then, we can use the fact that $(n-3)^2\ge 0$ for all $n$:
$n^2-6n+14 = (n-3)^2+5 \ge 5$
Therefore, $\frac{1}{n^2-6n+14}\leq \frac{1}{5n^2}$ for all $n$. 
Since $\sum_{n=1}^{\infty}\frac{1}{5n^2}$ converges by the comparison test with the p-series, we know that $\sum_{n=1}^{\infty}\frac{1}{n^2-6n+14}$ converges as well.

Determine Whether the Series Is Convergent or Divergent ∑n=1∞1n2−6n+14\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n ^ { 2 } - 6 n + 14 }∑n=1∞​n2−6n+141​

Determine Whether the Series Is Convergent or Divergent $\sum _ { n = 1 } ^ { \infty } \frac { 1 } { n ^ { 2 } - 6 n + 14 }$