Determine convergence or divergence of the series.
-$$\sum _ { n = 1 } ^ { \infty } \sin \left( \frac { 6 n ^ { 2 } + 5 } { n ^ { 2 } + 3 } \right)$$
A) Converges
B) Diverges

Question

Quizplus · Accepted Answer

We have $\lim_{n	o \infty} \sin\left(\frac{6n^2+5}{n^2+3}ight) = \sin(6)$, which is a constant. Therefore, by the limit test, the series $\sum_{n=1}^\infty \sin\left(\frac{6n^2+5}{n^2+3}ight)$ diverges.

Determine Convergence or Divergence of the Series ∑n=1∞sin⁡(6n2+5n2+3)\sum _ { n = 1 } ^ { \infty } \sin \left( \frac { 6 n ^ { 2 } + 5 } { n ^ { 2 } + 3 } \right)n=1∑∞​sin(n2+36n2+5​)

Determine Convergence or Divergence of the Series $\sum _ { n = 1 } ^ { \infty } \sin \left( \frac { 6 n ^ { 2 } + 5 } { n ^ { 2 } + 3 } \right)$