Mathematics 220 Workshop 3
1. Write down the following
(a) Definition of limn→∞ an = L
(b) Definition of a sequence {an } is bounded, an ∈ R.
P
(c) Definition of ∞
n=1 an converges.
2. Is the solution to each of the following question correct?
(a) ”For any x, y ∈ R with x < y + for any > 0, it holds x < y.” is false.
Solution: take x = 1, y = 0.5, = 1
Solution: The ”solution” is wrong. The chosen x, y do not satisfy x < y + for any
> 0.
(b) F/T and justify: If x, y 6∈ Q, then x + y 6∈ Q.
√
√
√
Proof. True. Take x = 2, y = 2, then x + y = 2 2 6∈ Q. Hence the statement is false.
Solution: The ”proof” is wrong. One cannot just check special values of x, y.
3. Show n5 ≤ 2n for all integer n ≥ 100.
Solution: We prove this by induction.
• Base case. When n = 100, 1005 = 210 · 510 and 2100 = 210 · 290 = 210 · (29 )10 . So
1005 < 2100 .
1
• Suppose that k 5 ≤ 2k for some k ≥ 100. Then note that 2 5 > 1, and so
1
1
(2 5 − 1)k ≥ (2 5 − 1)100 ≥ 1.
1
1 5
(To see the last inequality: it suffices to check 2 5 100 ≥ 101, or 2 ≥ (1 + 100
) .
5
5
2
3
4
5
Observe that (1+1/100) < (1+1/10) = 1+5/10+10/10 +10/10 +5/10 +1/10 =
1.61051 < 2) Thus,
1
25 k ≥ k + 1
and so
(k + 1)5 ≤ 2k 5 ≤ 2 · 2k = 2k+1 .
By mathematical induction, we have n5 ≤ 2n for all integer n ≥ 100.
4. Use two methods to show
∞
X
n3
n=1
2n
converges.
Solution:
Page 1 of ??
Mathematics 220 Workshop 3
• Method 1. By the previous question, for n ≥ 100,
n3
1
≤ 2.
n
2
n
P∞
P
5 n
2
and
Note that ∞
n=100 n /2 converges
n=1 1/n converges, so by comparison test,
P
5 /2n
adding finitely many (first 99 terms) does not affects convergence, so ∞
n
n=1
converges.
• Method 2.
an+1
(n + 1)3 2n
=
· n+1 =
an
n3
2
1 3 1
1
1+
· → < 1, as n → ∞.
n
2
2
Ratio test then implies the series is convergent.
5. Prove: If limn→∞ an = 0, then for any k > 0, limn→∞ akn = 0.
Solution: So suppose that k > 0 and an → 0. Choose any > 0. Then there exists an
1
N ∈ R so that n > N implies that |an − 0| < k . So then n > N implies that
1
|akn − 0| = |an |k < ( k )k = .
This shows that akn → 0.
6. Let {an } be a sequence of positive numbers. Show: limn→∞ an = ∞ iff limn→∞
1
an
= 0.
Solution: ”=⇒”. Giev any > 0, let M = 1/. Then there is a number N s.t. n > N
implies that an > M = 1/. Since each an is positive we have: for n > N
1
− 0 < .
an
Thus lim 1/an = 0.
”⇐=”. So suppose that an is a sequence of positive numbers with a1n → 0. To show
an → ∞, choose any M ∈ R. Since an > 0 for each n, then if M ≤ 0 then we automatically
have an > M for all n ∈ N. So suppose that M > 0.
Then a1n → 0 implies that there is a number N so that n > N implies that
Hence, n > N implies that M < an , which shows that an → ∞.
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1
an
<
1
M.