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

1 Answer
Apr 25, 2018

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[1,),1sin(2n)1.

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

b_n=1/(1+2^n)bn=11+2n

We can also drop the constant 11 from the denominator. This will not drastically change the behavior of b_nbn in terms of it being larger than a_nan:

b_n=1/2^n=(1/2)^nbn=12n=(12)n

Now, we know the series

sum_(n=1)^oo(1/2)^nn=1(12)n converges as it is a geometric series with the common ratio |r|=1/2<1|r|=12<1.

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