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Quiz 3 Key MA 366 Problem 1: Give a careful proof that if ak = k 2 /(k 2 + 1) then limk ak = 1, USING THE DEFINITION OF CONVERGENCE. Scratchwork: There are many ways to proceed. We want |k 2 /(k 2 +1)−1| < ϵ, which simplifies to | − 1/(k 2 + 1)| < ϵ or 1 < ϵ. +1 Reciprocate, subtract 1 from both sides, take a square root and get k > √ 1/ϵ − 1. We can take N as the right side, or to be on the√safe side forget the −1, round up to the nearest integer, and take N = ⌈1/ ϵ⌉; note N ≥ 1 always. k2 √ Polished Proof: Given ϵ > 0 let N = ⌈1/ ϵ⌉. If k ≥ N then √ k ≥ ⌈1/ ϵ⌉ √ ≥ 1/ ϵ so that 1/k 2 ≤ ϵ. Thus 2 k |ak − 1| = 2 − 1 k +1 1 k2 + 1 1 < 2 k ≤ ϵ. ∑ Problem 2: Suppose the series ak converges. Prove that the series ∑ ak 2k k = converges. ∑ Proof: Probably a lot of ways to do this. First, if ak converges then limk ak = 0. Thus the sequence |ak | is bounded (e.g., for some N we have |ak | ≤ 1 for all k ≥ N , so all |ak | are bounded above by M = max(|a1 |, . . . , |aN −1 |, 1)). We then have M |ak | ≤ k. k 2 2 The series∑ with terms on the right converges (it’s geometric, ratio 1/2), so ∑ the series k |ak |/2k converges and so k ak /2k converges (absolutely). 1