# How do you test for convergence (sin(2n))/(1+(2^n)) from n=1 to infinity?

Apr 25, 2018

Converges by the Direct Comparison Test.

#### Explanation:

We can use the Direct Comparison Test for this.

On the interval $\left[1 , \infty\right) , - 1 \le \sin \left(2 n\right) \le 1$.

So, for our comparison sequence ${b}_{n} ,$ if we remove $\sin \left(2 n\right)$ from the denominator, we get a larger numerator and therefore a larger sequence:

${b}_{n} = \frac{1}{1 + {2}^{n}}$

We can also drop the constant $1$ from the denominator. This will not drastically change the behavior of ${b}_{n}$ in terms of it being larger than ${a}_{n}$:

${b}_{n} = \frac{1}{2} ^ n = {\left(\frac{1}{2}\right)}^{n}$

Now, we know the series

${\sum}_{n = 1}^{\infty} {\left(\frac{1}{2}\right)}^{n}$ converges as it is a geometric series with the common ratio $| r | = \frac{1}{2} < 1$.

Then, since the larger series converges, so must the smaller series.