test the series for convergence or divergence. \n∑(n = 1 to ∞) sin(6n)/(1 + 3^n)\nconvergent\ndivergent\nneed…

Explanation:

Step1: Apply comparison test

We know that (|\sin(6n)|\leq1) for all (n). So, (\left|\frac{\sin(6n)}{1 + 3^{n}}\right|\leq\frac{1}{1 + 3^{n}}\leq\frac{1}{3^{n}}).

Step2: Consider geometric - series

The series (\sum_{n = 1}^{\infty}\frac{1}{3^{n}}) is a geometric series with common ratio (r=\frac{1}{3}). For a geometric series (\sum_{n = 1}^{\infty}ar^{n - 1}) (in our case (a=\frac{1}{3}), (r=\frac{1}{3})), and since (|r|=\left|\frac{1}{3}\right|<1), the geometric series (\sum_{n = 1}^{\infty}\frac{1}{3^{n}}) converges.

Step3: Use absolute - convergence implies convergence

By the comparison test, since (\sum_{n = 1}^{\infty}\left|\frac{\sin(6n)}{1 + 3^{n}}\right|\leq\sum_{n = 1}^{\infty}\frac{1}{3^{n}}) and (\sum_{n = 1}^{\infty}\frac{1}{3^{n}}) converges, the series (\sum_{n = 1}^{\infty}\frac{\sin(6n)}{1 + 3^{n}}) is absolutely convergent. And absolute - convergence implies convergence.

Answer:

convergent