Final answer:
The series ∑ (-1)^ⁿ¹ * sin²(1/n) converges absolutely because it meets the criteria of the Alternating Series Test and its absolute series converges, resembling the behavior of a convergent p-series for large n.
Step-by-step explanation:
To determine if the series ∑ from n=1 to ∞ of (-1)^ⁿ¹ * sin²(1/n) converges conditionally or absolutely, we should first acknowledge that this is an alternating series due to the (-1)^ⁿ¹ term. To apply the Alternating Series Test, we check if the absolute value of the terms decreases monotonically to zero. Notice that as n increases, sin²(1/n) approaches zero since sin²(x) approaches zero as x approaches zero.
For absolute convergence, we consider the absolute value of the series terms without the (-1)^ⁿ¹ factor, which is the series of sin²(1/n). We need the absolute terms to converge as well. Knowing that sin(x) ≈ x for small x, sin²(1/n) approximately equals (1/n)². Hence, the comparison test can be used by comparing our series to the p-series ∑ 1/n², which is convergent for p>1.
Since the original alternating series passes the Alternating Series Test and the absolute series converges (as it behaves like a convergent p-series for large n), we can conclude that the given series converges absolutely.