Macroeconomics Exam Review 58 - Solutions for Foundations...

since the term in brackets is strictly greater than 1 for any 𝑥 > 0. Similarly
𝑒
𝑥
𝑥
=
( 𝑒
𝑥/ ( 𝑛 +1)
)
𝑛
𝑒
𝑥/ ( 𝑛 +1)
( 𝑛 + 1) 𝑛 (
𝑥
𝑛 +1
) 𝑛
=
1
( 𝑛 + 1) 𝑛
(
𝑒 𝑥/ ( 𝑛 +1)
𝑥/ ( 𝑛 + 1)
)
𝑛
𝑒
𝑥/ ( 𝑛 +1)
→ ∞
2.8 Assume that 𝑆 ⊆ ℜ is compact. Then 𝑆 is bounded (Proposition 1.1), and there
exists 𝑀 such that ∣ 𝑥 ∣ ≤ 𝑀 for every 𝑥 ∈ 𝑆 . For all 𝑛 ≥ 𝑚 ≥ 2 𝑀
∣ 𝑓
𝑛
( 𝑥 ) − 𝑓
𝑚
( 𝑥 ) ∣ =
𝑛 ∑
𝑘 = 𝑚 +1
𝑥
𝑘
𝑘 !
≤
𝑥
𝑚 +1
( 𝑚 + 1)!
𝑛 − 𝑚 ∑
𝑘 =0
(
𝑥
𝑚
)
𝑘
≤
𝑀 𝑚 +1
( 𝑚 + 1)!
𝑛 − 𝑚 ∑
𝑘 =0
(
𝑀
𝑚
)
𝑘
≤
𝑀 𝑚 +1
( 𝑚 + 1)!
(
1 +
1
2
+
1
4
+ ⋅ ⋅ ⋅ +
(
1
2
)
𝑛 − 𝑚
)
≤ 2
𝑀
𝑚 +1
( 𝑚 + 1)!
≤ 2
(
𝑀
𝑚 + 1
)
𝑚 +1
≤
(
1
2
)
𝑚
by Exercise 1.206. Therefore 𝑓
𝑛
converges to 𝑓 for all 𝑥 ∈ 𝑆 .
2.9 This is a special case of Example 2.8. For any 𝑓, 𝑔 ∈ 𝐹 ( 𝑋 ), define
( 𝑓 + 𝑔 ) = 𝑓 ( 𝑥 ) + 𝑔 ( 𝑥 )
( 𝛼𝑓 )( 𝑥 ) = 𝛼𝑓 ( 𝑥 )
With these definitions 𝑓 + 𝑔 and