craig16 said:So , 1/n is always greater than sin(1/n)
If I just replace sin(1/n) with 1/n , I have a "an" that is less than the original and the new function diverges by the integral test.
So the sum is divergent by the comparison test?
craig16 said:Okay, I see how the test would work replacing sin(1/n) with 1/2n making the new equation always less than the original. Then the integral test works and the sum is divergent.
Does that mean that wolfram is wrong though?
http://www.wolframalpha.com/input/?i=sum from 2 to inf (sin(1/n))/(sqrt(ln(n)))&t=macw01
When I plug the sum in it tells me it converges to 962 or so.
Convergence tests for series are used to determine whether an infinite series, which is a sum of infinitely many terms, converges to a finite value or not. This helps us determine the behavior and properties of the series and whether it can be used in calculations or not.
Absolute convergence means that a series converges to a finite value regardless of the order in which the terms are added. Conditional convergence means that a series only converges if the terms are added in a specific order. For example, the alternating harmonic series is conditionally convergent, while the regular harmonic series is absolutely convergent.
The most commonly used test is the ratio test, which compares the ratio of consecutive terms in a series to a limit to determine whether the series converges or diverges.
The convergence of a power series can be determined by using the root test, which compares the nth root of the absolute value of the series to a limit to determine convergence or divergence. If the limit is less than 1, the series converges, and if it is greater than 1, the series diverges.
No, a series can only converge to one value. If a series converges, it means that the sum of its terms approaches a finite value as the number of terms increases. If a series were to converge to more than one value, it would imply that the sum of its terms is both finite and infinite, which is not possible.