hivesaeed4 said: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?
Determining convergence of a series helps us understand whether the series will approach a finite value or diverge to infinity as the number of terms increases. This can be useful in various fields of science, including physics, engineering, and economics.
The general method for determining convergence of a series is to evaluate the limit of the series as the number of terms approaches infinity. If the limit is a finite value, then the series is said to converge. If the limit is infinity or does not exist, then the series is said to diverge.
To determine the convergence of ((sin(n))^4)/(1+n^2), you can use the comparison test or the limit comparison test. You can also use the ratio test or the root test, although these may be more complicated in this case. It is also important to check for any special cases, such as when n=0, to ensure the tests are applicable.
The comparison test states that if 0 ≤ a_n ≤ b_n for all n ≥ N, where N is some positive integer, and b_n converges, then a_n also converges. In the case of ((sin(n))^4)/(1+n^2), we can compare it to the series 1/(n^2), which is known to converge. As 0 ≤ ((sin(n))^4)/(1+n^2) ≤ 1/(n^2) for all n ≥ 1, we can conclude that ((sin(n))^4)/(1+n^2) also converges.
If ((sin(n))^4)/(1+n^2) converges, it means that as the number of terms increases, the series approaches a finite value. In this case, the series will approach 0 as n approaches infinity. This can be interpreted as the series having a sum of 0, or the series becoming infinitely small. This information can be useful in various scientific applications.