determine if the series is absolutely convergent and convergent
the sum from n=1 to infinity of
sin(n^2)/n^2
what series test should I use and how?
the ratio test?
|sin(n^2)/n^2 | < 1/n^2 and sum from 1 to infinity of 1/n^2 converges (to pi^2/6 )
To determine if the series is absolutely convergent and convergent, you can use the ratio test. The ratio test is a common method for testing the convergence of infinite series.
To apply the ratio test to the given series:
1. Take the absolute value of each term in the series. The absolute value of sin(n^2)/n^2 is |sin(n^2)/n^2|.
2. Compute the limit of the absolute value of the ratio of consecutive terms as n approaches infinity:
lim(n→∞) (|sin((n+1)^2)/(n+1)^2|) / (|sin(n^2)/n^2|)
3. Simplify the expression.
lim(n→∞) |sin((n+1)^2)/((n+1)^2)| * |n^2/sin(n^2)|
4. Apply the trigonometric identity: sin(x)/x → 1, as x → 0.
lim(n→∞) |(n+1)^2/n^2| * |1| = lim(n→∞) (1 + 2/n + 1/n^2) = 1
5. Examine the result. If the limit is less than 1, the series is absolutely convergent. If the limit is greater than 1 or equal to infinity, the series is divergent. If the limit is exactly 1, the test is inconclusive.
In this case, the series you provided has a limit of 1. Since the limit is equal to 1, the ratio test is inconclusive, and you cannot determine the absolute convergence or divergence of the series using the ratio test alone. Other tests such as the comparison test or the integral test may be used to provide more information about the convergence or divergence of the series.