... *Similar to 4.1.6, but with different numbers n + 5n Given any ε > 0, we want to make –0 < ε n –7 By making n ≥ we can remove the absolute value signs since n – will be positive n + 5n So we want to ... ∈ and some m ∈ If for some k > we have |sn – s| ≤ k | |, an for all n ≥ m, and if lim an = 0, then it follows that lim sn = s Proof: Given any ε implies that | an | < ε /k > 0, since lim an ... denominator 2 For large values of n, the numerator behaves like n , so we want n + 5n ≤ bn 3 And the denominator behaves like n , so we want n – ≥ cn Then we will have n + 5n n –7 ≤ = bn cn b c n and...