Question

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

Answer 1

Converges by the Direct Comparison Test.

Explanation:

We can use the Direct Comparison Test for this.

On the interval `[1, oo), -1<=sin(2n)<=1`.

So, for our comparison sequence `b_n,` if we remove `sin(2n)` from the denominator, we get a larger numerator and therefore a larger sequence:

`b_n=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=1/2^n=(1/2)^n`

Now, we know the series

`sum_(n=1)^oo(1/2)^n` converges as it is a geometric series with the common ratio `|r|=1/2<1`.

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