# Determining Convergence of ((sin(n))^4)/(1+n^2)

1. Apr 2, 2012

### hivesaeed4

Suppose I want to determine the convergence of ((sin(n))^4)/(1+n^2) using limit comparison test. I divide it by 1/(1+n^2). All that remains is (sin(n))^4. Now as the limit goes to infinty, the range of values (sin(n))^4 can give is 0 to 1. Now it gives many more values above zero then at zero, so is that why we declare it a convergent series?

2. Apr 3, 2012

### M Quack

As you point out, (sin(n))^4 is always between 0 and 1.

therefore

0 <= ((sin(n))^4)/(1+n^2) < 1/(1+n^2) < 1/n^2 for all n>0.

1/n^2 converges. So we're home and dry.

3. Apr 3, 2012

### DonAntonio

Well, you almost have it : $\frac{\sin^4(n)}{1+n^2}\leq\frac{1}{n^2}$ , which is clearly convergent.

DonAntonio

4. Apr 3, 2012

### hivesaeed4

Lots of Thanks.