# Quiz 3 Key ```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| &lt;
ϵ, which simplifies to | − 1/(k 2 + 1)| &lt; ϵ or
1
&lt; ϵ.
+1
Reciprocate, subtract 1 from both sides, take a square root and get k &gt;
√
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 &gt; 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
&lt; 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
```