AP Calculus BC - Series Convergent/ Divergent

Question

1. 1 to infinity, sin(1/x^2)
2. 1 to infinity, 1/n(n+2)(n+5)
3. 1 to infinity, ((-1)^n n^8 + 15)/ n^8 + 14

Answer 1

1. I'm assuming the series term is sin(1/n^2). This converges because it is bounded from above by 1/n^2, which is a convergent series with p = 2. Use comparison or limit comparison test with 1/n^2.

2. Converges. Bounded from above by 1/n^3, which is a convergent series with p =3. Use comparison or limit comparison test with 1/n^3.

3. Diverges. The given term doesn't go to zero as n -> infinity.

Note: you must put the full denominator in parentheses, or it means something else: a / b + c is not the same as a / (b+c)