Guest Session: 1 Question Remaining. Create Account to save progress.
Login
Infinite Seriesmedium
0:00.0

Consider ∑n=1∞anbn\sum_{n=1}^{\infty} \frac{a_n}{b_n}∑n=1∞​bn​an​​ where an=sin⁡2(1/n)a_n = \sin^2(1/n)an​=sin2(1/n) and bn=1/n2b_n = 1/n^2bn​=1/n2. What is the conclusion by the Limit Comparison Test?