Series divergence and convergence question

Question

Suppose 0 ≤ b_n ≤ 2ⁿ ≤ a_n and 0 ≤ c_n ≤ 2^-n ≤ d_n for all n. Of the series, ∑a_n , ∑b_{n ,} ∑c_n, ∑d_n, which will definitely converge and which will definitely diverge? Explain.

Answer 1

C_n definitely converges because it's less than or equal to the 2^-n which means 1/2ⁿ which is a geometric series with r = 1/2, and geometric series converge when the r is between -1 and 1 not including -1 and 1. C_n sum is less than one that converges then it also converges. 2^-n converges to 1 or 2 depending on whether it starts at n=0 or n=1 so C_n is going to be less than 1 or 2.

d_n we don't know for sure because it's greater than or equal to the 2^-n series which converges to 1 or 2, so it could be 3, 4, or 5 or infinity, all we know is it's greater than or equal to 1 or 2

b_n we don't know for sure because it's less than or equal 2ⁿ which diverges (geometric r = 2 which is > 1)to infinity so we just know its less than that but it could still diverge to infinity, saying it's less than the series that adds to infinity doesn't mean it converges because it could be infinity - 1 which is still infinity.

a_n definitely diverges because it's greater than or equal to the divergent series 2ⁿ

Answer 2

The sum of a_n diverges.

The sum of b_n is indeterminate.

The sum of c_n converges.

The sum of d_n is indeterminate.

I will leave the explanation part to you.