Can the Alternating Series Test be applied to ∑n=1∞(−1)nsin(π2n)\sum_{n=1}^{\infty} (-1)^n \sin\left(\frac{\pi}{2n}\right)∑n=1∞(−1)nsin(2nπ)?
Yes, because sin(π/(2n))→0\sin(\pi/(2n)) \to 0sin(π/(2n))→0 and the terms are decreasing.
No, because sin(π/(2n))\sin(\pi/(2n))sin(π/(2n)) is not strictly monotone decreasing.
Yes, because the limit of terms is zero, regardless of monotonicity.
No, because the terms do not approach zero fast enough.