MATH 444, Spring 2011

Assignment #4 - Solution

1. Show lim (n2−1)(2n2+3)

= 12.

Discussion: We want to show:

∀ε > 0, there exists a K ∈ N such that for all n ≥ K∣∣∣∣ (n2 − 1)

(2n2 + 3)− 1

2

∣∣∣∣ =

∣∣∣∣2(n2 − 1)− (2n2 + 3)

2(2n2 + 3)

∣∣∣∣ =5

4n2 + 6<

8

4n2<

2

n< ε.

So we can take K > 2ε.

Proof: Let ε > 0, then there exists K ∈ N such that K > 2ε. Then for all n ≥ K we

have ∣∣∣∣ (n2 − 1)

(2n2 + 3)− 1

2

∣∣∣∣ =

∣∣∣∣2(n2 − 1)− (2n2 + 3)

2(2n2 + 3)

∣∣∣∣ =5

4n2 + 6<

8

4n2<

2

n< ε.

2. Prove that if limxn = x > 0, then there exists a natural number M such that xn > 0for all n ≥M .

Proof : Let ε = x2> 0. Since limxn = x, there exists a positive integer M such that

for all n ≥M , |xn − x| < ε. In this case, we have

0 <x

2< x− ε < xn.

3. Let x1 = 1 and xn+1 =√

2 + xn for all n ∈ N. Show that (xn) converges and find thelimit.

Proof: (1) Use math induction to show that 0 ≤ xn ≤ 2.

(2) Show (xn) is monotone increasing.

For instance, use induction. It is clear that x1 = 1 < x2 =√

2 + 1 =√

3. Suppose thatx1 < x2 < · · ·xk+1. Then we have xk+1 =

√xk + 2 < xk+2 =

√xk+1 + 2. So (xn) is a

strictly increasing sequence.

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(3) Let x = limxn. Then we get x =√x+ 2. It follows that x2 − x + 2 = 0, and we

have limxn = 2.

4. Let (xn) be a bounded sequence and let s = sup{xn : n ∈ N}. If s /∈ {xn : n ∈ N},show that there exists a subsequence (xnk

) that converges to s.

Proof: We need an induction procedure.

For n = 1, since s− 1 < s, there exists an element, say xn1 such that

s− 1 < xn1 < s.

Suppose that for n = k, we have chosen xn1 , · · · , xnksuch that

s− 1

j< xnj

< s

for j = 1, 2, · · · , k and n1 < n2 < · · · < nk.

Now we consider n = k + 1. Since xn 6= s for all n = 1, 2, · · · , nk, we get max{s −1

k+1, x1, x2, · · · , xnk

} < s. Then there exists an element, say xnk+1such that

s− 1

k + 1≤ max{s− 1

k + 1, x1, x2, · · · , xnk

} < xnk+1≤ s.

In this case, it is clear that nk+1 > nk. Therefore, by math induction, we get an(momotone increasing) subsequence (xnk

) such that

s− 1

k< xnk

≤ s.

Therefore, (xnk) converges to s.

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Extra Problems for 4 Hours Credit Students

5. Let yn =√n+ 1−

√n. Show that yn and

√nyn converge. Find their limits.

Proof: We first note that

0 ≤√n+ 1−

√n =

(√n+ 1−

√n)(√n+ 1 +

√n)√

n+ 1 +√n

=1√

n+ 1 +√n≤ 1√

n→ 0,

we can conclude that lim yn = 0.

We also have

√n(√n+ 1−

√n) =

√n

(√n+ 1−

√n)(√n+ 1 +

√n)√

n+ 1 +√n

=

√n√

n+ 1 +√n≤

√n√

n+√n

=1

2,

and

√n(√n+ 1−

√n) =

√n

(√n+ 1−

√n)(√n+ 1 +

√n)√

n+ 1 +√n

=

√n√

n+ 1 +√n≥

√n√

n+ 1 +√n+ 1

=

√n

2√n+ 1

Since lim nn+1

= lim(1− 1n+1

) = 1, we have lim√n

2√n+1

= 12, and so we can also conlude

that lim√nyn = 1

2.

6. Let (xn) be an unbounded sequence. Show that there exists a subsequence (xnk) such

that lim 1xnk

= 0.

Proof: We first note that since (xnk) is an unbounded sequence, for any k ∈ N, there

exists an element xnkin the set {xn} such that |xnk

| > k. However, this argumentcan not guarantee such selected {xnk

} satisfies n1 < n2 < · · ·nk · · · To construct asubsequence, we need to use the following induction method.

First for n = 1, there exists an element, say xn1 , such that |xn1| > 1.

Then we see that the remainder sequence (xn1+1, · · · , xn+m · · · ) is also unbounded.For n = 2, there exists an element, say xn2 from this remainder sequence such that|xn2| > 2. It is clear that we have n1 < n2.

Now suppose that for n = k, we have chosen

|xn1 | > 1, · · · , |xnk| > k

3

and n1 < n2 < · · · < nk.

Then the remainder sequence (xnk+1, · · · , xnm , · · · ) is again unbounded, and we canchoose an element xnk+1

from this remainder sequence such that∣∣xnk+1

∣∣ > k + 1.

Then by math induction, we can get a subsequence (xnk) such that

|xnk| > k

for all k ∈ N. Therefore, we have 0 <∣∣∣ 1xnk

∣∣∣ < 1k→ 0.

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