Convergence to 0

Question

Prove that the sequence is convergent to 0, using monotonic convergence:

a) (n^2)/(2^n)

b) (10^n)/((2n+1)!)

Answer 1

a) n²/2ⁿ≥0 for all n, and n²/2ⁿ=0 only when n=0. When n=1 n²/2ⁿ=½. n=0 establishes the infimum (lower bound) for the series. The monotonic convergence theorem states that if a series has progressively decreasing terms, and is bounded below it will converge to the lower bound. So the series converges to zero.

b) When n=0 10ⁿ/(2n+1)!=1/1=1. When n=1 it is 10/6=5/3; when n=2 it's 100/120=⅚. The series is decreasing so the lower bound is 1 (at n=0) and the series convergences to 1 according to the theorem.