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

[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?

 

[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.

 

[DonAntonio]

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.


*** Perhaps you meant "...can give is BETWEEN 0 and 1"...? ****


Now it gives many more values above zero then at zero, so is that why we declare it a convergent series?

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

DonAntonio

 

[hivesaeed4]

Lots of Thanks.