suppose that ( a_{n}=sin (n pi / 2) / n ). find the smallest integer ( n ) with the property that (…

Explanation:

Step1: Analyze the general term

Given (a_{n}=\frac{\sin(n\pi/2)}{n}), we know that (|\sin(n\pi/2)|\leq1). So (|a_{n}|=\left|\frac{\sin(n\pi/2)}{n}\right|\leq\frac{1}{n}).

Step2: Solve the inequality

We want (|a_{n}|<\frac{1}{10}). Since (|a_{n}|\leq\frac{1}{n}), we solve the inequality (\frac{1}{n}<\frac{1}{10}). Cross - multiplying gives (n > 10).

Step3: Check specific values

Let's check some values of (n). When (n = 11), (\sin\left(\frac{11\pi}{2}\right)=\sin\left(5\pi+\frac{\pi}{2}\right)=-\sin\left(\frac{\pi}{2}\right)=- 1), and (|a_{11}|=\left|\frac{\sin(11\pi/2)}{11}\right|=\frac{1}{11}<\frac{1}{10}). When (n = 10), (\sin\left(\frac{10\pi}{2}\right)=\sin(5\pi) = 0), and (|a_{10}|=\left|\frac{\sin(10\pi/2)}{10}\right| = 0<\frac{1}{10}). But we need to check the worst - case scenario (when (|\sin(n\pi/2)| = 1)). For (n = 9), (\sin\left(\frac{9\pi}{2}\right)=\sin\left(4\pi+\frac{\pi}{2}\right)=\sin\left(\frac{\pi}{2}\right)=1), and (|a_{9}|=\left|\frac{\sin(9\pi/2)}{9}\right|=\frac{1}{9}>\frac{1}{10}).

Answer:

(N = 11)