Epsilon Ics 1 Sol

7/23/2019 Epsilon Ics 1 Sol

http://slidepdf.com/reader/full/epsilon-ics-1-sol 1/6

Math 347 Worksheet: Epsilonics, I: Sequences and Limits —SOLUTIONS— A.J. Hildebrand

Practice Problems Solutions

The problems below have been carefully selected to illustrate common situations and the techniques and tricks to deal withthese. Try to master them all; it is well worth it! What you learn in the process will be useful later in this

class when we get to epsilonics in other contexts, and in many advanced math classes.

For each problem, first try to gain an intuitive “feel” for the problem (a sketch of a typical situation may be useful) andtry to understand why the statement is correct. Then try to construct a rigorous -proof.

In the following {an}, {bn}, and {cn} denote arbitrary sequences, and L and M denote real numbers. Unless otherwisespecified, you should only use the -definition of a limit, not any theorems or properties of limits.

1. Warmup problems. These are conceptually quite easy, and the results are intuitively “clear”. Try to use theseproblems to practice proper write-ups of proofs.

(a) Limit of constant sequence. Prove that the limit of a constant sequence is equal to this constant; i.e., showrigorously that, if an = c for all n, then lim

n→∞an = c.

Solution:

Proof: By the definition of a limit, we need to show that, given any > 0, there exists an N ∈ N such that, for all n ≥ N , (∗) |an − c| < . Now, if an = c for all n ∈ N, then |an − c| = 0 for all n ∈ N, so (∗) holds for any > 0and all n ∈ N. Thus, the definition of a limit is satisfied with N = 1.

(b) Scaling property.

Prove that if limn→∞an

= L, then for any c ∈R

, limn→∞(can

) = cL.Solution: Idea and proof strategy. We use a variation of the /2 trick: By the definition of a limit, we have(∗) |an − L| < from some point onwards. Multiplying (∗) by |c|, gives (∗∗) |can − cL| < |c|, which is an estimateof the type we want except that we need instead of |c| on the right. We can achieve this if we start out in (∗)with = /(|c| + 1) in place of . (Note that using /|c| would cause problems in case c = 0; Replacing |c| in thedenominator by |c| + 1 avoids this issue.)Below is a formal proof along these lines..

Proof: Suppose limn→∞

an = L, and let c be a given real number. Applying the definition of a limit with

= /(|c| + 1), we obtain an N ∈ N such that

(1) |an − L| < =

|c| + 1 for all n ≥ N .

Then for n ≥ N we have

|can − cL| = |c| · |an − L| (by properties of absolute values)

< |c|

|c| + 1 (by (1))

< .

Hence n ≥ N implies |can − cL| < .By the definition of a limit, this proves that lim

n→∞

(can) = cL.

(c) Multiplication by bounded sequence. Prove that if limn→∞

an = 0 and {bn} is a bounded sequence, then

limn→∞

anbn = 0.

Solution: Idea and proof strategy. We need to get from an inequality of the form (∗) |an − 0| < to aninequality of the form (∗∗) |anbn − 0| < . Since the sequence bn is bounded, we have (with a suitable bound M )

|bn| ≤ M for all n, so (∗) implies |anbn − 0| ≤ |M | · |an| < |M |. To get a bound < in the last step, we replace in (∗) by = /(|M | + 1). (Note again, the trick of working with |M | + 1 in place of |M | in order to ensure that is well-defined.)Now the formal proof:

Proof: Let > 0 be given. Since limn→∞

an = 0, applying the definition of a limit with = /(|M | + 1). we obtain

an N ∈ N such that|an| = |an − 0| <

|M | + 1 for all n ≥ N .

Then|anbn − 0| = |anbn| < |an||bn| <

|M | + 1 · |M | < for all n ≥ N .

Hence n ≥ N implies |anbn − 0| < .By the definition of a limit, this proves that lim

n→∞

can = cL.

1




(d) Limit of shifted sequence. Let an be a given sequence, and let bn = an+1 be the “shifted” sequence. Provethat if lim

n→∞an = L, then lim

n→∞bn = L.

Solution: Proof sketch. We need to get from an inequality of the form (∗) |an − L| < to an inequality of theform (∗∗) |an+1 − L| < . Now if (∗) holds for all n ≥ N , then (∗∗) holds for all n ≥ N − 1 and therefore also forall n ≥ N . Thus the N -value that “worked” for the original sequence also “works” for the shifted sequence.

2. Intermediate problems. These problems illustrate a variety of techniques and tricks in working with limits.

(a) Uniqueness of limit. Prove that if limn→∞

an = L and limn→∞

an = M , then L = M . (Hint: Use contradiction.)

Solution: Idea and proof strategy. Given the form of the claim, the most natural approach is a proof bycontradiction. Thus, we assume {an} has two distinct limits, L and M , and seek to derive a contradiction fromthis assumption. Applying the limit definition with both L and M , we see that, given any > 0, from some pointonwards all terms an must fall into the intervals (∗) L − < an < L + and (∗∗) M − < an < M + . We wilget a contradiction if we choose small enough so that these two intervals don’t overlap, for example, by choosing = |M − L|/3.

Proof: We argue by contradiction. Suppose limn→∞

an = L, limn→∞

an = M , and L = M . Then either L < M or

L > M . Without loss of generality, we may suppose L < M . (Otherwise we could just relabel the limits, setting L = M and M = L and working with L and M .)Applying the definition of a limit with = (M − L)/3 (which is positive by our assumption L < M ), we obtain

N 1, N 2 ∈ N such that

|L − an| < for all n ≥ N 1,(1)

|M − an| < (2)

Let N = max(N 1, N 2). Then aN satisfies both (1) and (2), so in particular,

aN < L + (by (1)) and aN > M − (by (2)).

This implies M − < L + , and hence, by the definition of ,

M − 1

3(M − L) < L +

1

3(M − L),

2

3(M − L) < 0,

which is a contradiction to our assumption L < M .Hence we must have L = M .

Remark: Note the “Without loss of generality” phrase used at the beginning of the argument. This is a verycommon phrase (often abbreviated as WLOG) used to add assumptions that are convenient in the proof, butwhich do not alter the nature of the problem in any material way. Here the additional assumption is L < M Since the cases L < M and L > M are completely symmetrical (relabeling M as L and L as M converts onecase into the other), the “WLOG” assumption is entirely justified here.

(b) Preservation of inequalities. Prove that if limn→∞

an = L, limn→∞

bn = M , and an < bn for all n ∈ N, then L ≤ M

Solution: Proof idea. This is similar to the proof that the limit of a sequence is unique. Argue by contradictionassume L < M , and apply the definition of convergence with = (M − L)/3.

(c) Squeeze Theorem. Suppose an ≤ bn ≤ cn for all n ∈ N. Show that if limn→∞


cn = L, then

limn→∞

bn = L.

Solution: Idea and proof strategy. Since an and cn both have limits L, we know that |an − L| < and|cn − L| < hold from some point onwards. We write these bounds as two-sided inequalities for an and cnL − < an < L + , and L − < cn < L + Since bn ≥ an, the first of these inequalities implies bn ≥ an > L − similarly, since bn ≤ cn, the second relation implies bn < L + . Altogether we get L − < bn < L + , i.e.|bn − L| < , from some point onwards.

(d) Cauchy Criterium, “easy” direction. Prove that any convergent sequence is a Cauchy sequence. (Hint: Usethe /2-trick.)

Solution: Idea and proof strategy. Comparing the definitions of “convergent sequence” and “Cauchy se-quence”, we see that we need to convert a bound of the form ( ∗) |an − L| < to a bound of the form (∗∗)

2




|am − an| < . . The trick that accomplishes this is a standard one: Add and subtract L in am − an, then applythe triangle inequality:

|am − an| = |(am − L) + (L − an)| ≤ |am − L| + |an − L|.

Combining this with another standard trick, namely working with = /2 in place of in (∗), gives the desired-bound.Here is a formal proof:

Proof: Suppose limn→∞an = L. Let > 0 be given. Since limn→∞an = L, applying the definition of a limit with = /2, we obtain N ∈ N such that

(1) |an − L| <

2 for all n ≥ N .

Let m, n ≥ N . Then

|an − am| = |(an − L) + (L − am)| (by algebra)

≤ |an − L| + |L − am| (by the triangle inequality)

<

2 +

2 (by (1) applied to am and an)

= .

Hence m, n ≥ N implies |an − am| < .

Thus the sequence {an} satisfies the Cauchy Criterion, and hence is a Cauchy sequence.

(e) Convergent implies bounded. Prove that any convergent sequence is bounded. (Hint: Apply the definition oa limit with = 1.)

Solution: Idea and proof strategy. To show that a sequence {an} is bounded, we we need to show that thereexists an M such that |an| ≤ M for all n. We proceed in two stages.

First, applying the definition of a limit with = 1, we obtain such a bound (namely, |L| + 1, where L is the limitof the sequence) that holds for all terms an with n greater than some suitable N .

To extend this to a bound for all an, we need to deal with the terms an for n ≤ N . We cannot say anythingabout these terms, but since there are only finitely many such terms, they have a finite upper bound (namely, thelargest among the N terms |a1|, . . . , |aN |).

Combining these two bounds gives the required global bound for an.Here is a formal proof:

Proof: Let {an} be a convergent sequence and let L be its limit. Applying the definition of a limit with = 1,we obtain an N ∈ N such that

(1) |an − L| < 1 for all n ≥ N .

Set

M 1 = |L| + 1,(2)

M 2 = max{|a1|, |a2|, . . . , |aN |},(3)

M = max(M 1, M 2).(4)

We will show that, with this choice of M , we have |an| ≤ M for all n ∈ N.If n ≥ N , then

|an| = |(an − L) + L| (by algebra)≤ |an − L| + |L| (by triangle inequality)

≤ 1 + |L| (by (1) since n ≥ N )

= M 1 (by def. (2) of M 1)

≤ M (by def. (4) of M ).

If n ≤ N , then

|an| ≤ max{|a1|, |a2|, . . . , |aN |}

= M 2 (by (def. (3) of M 2)

≤ M (by def. (4) of M ).

Thus, |an| ≤ M holds for all n ≤ N .Therefore the sequence {an} is bounded.

3




Remarks: Note that we needed the definition of a limit of {an} only for a single value of ; we chose = 1though any other positive number would have worked. This was enough to yield the boundedness of the sequence

Once N has been chosen, the maximum in (3) is a maximum over a finite set of real numbers and hence is well-defined. That a finite set of real numbers has a maximum, i.e., that one of these numbers must be greater or equalto all others, does not need to be justified, but could be proved formally by induction on the number of terms,using the fact that, for any two real numbers x and y we have either x ≤ y (and so y = max(x, y)), or x > y (andso x = max(x, y)). (This fact is one of the axioms defining the real numbers, so does not need a proof.)

(f) Cauchy implies bounded. Prove that any Cauchy sequence is bounded.

Solution: Idea and proof strategy. We use the same idea as in the previous problem, showing first that thesequence is bounded in a range n ≥ N with a suitable N , then extending this bound to one for the full rangen ∈ N. For the first part of the argument, we apply the definition of a Cauchy sequence with = 1 to obtain anN ∈ N such that |an − am| < 1 holds for all n, m ≥ N . Applying this with m = N , we get |an − aN | < 1 for aln ≥ N , and via the triangle inequality this gives the bound |an| ≤ |aN | + 1 for all n ≥ N . Thus, M 1 = |aN | + 1is a bound for an in the range n ≥ N . (Note that in aN the index N is a fixed index that has been chosen at thebeginning of the proof, namely, the N that is obtained when applying the definition of a Cauchy sequence with = 1; thus, aN is also a fixed number, and it is okay to define M 1 in terms of aN as was done above.)

Using the same argument as in the previous problem, one can extend this bound to one that applies to the fulrange n ∈ N,

3. Harder problems. The following two problems are somewhat trickier, but very instructive.

(a) Reciprocal property. Prove that if limn→∞

an = L, an = 0 for all n ∈ N, and L = 0, then limn→∞

1

an= 1

L. (Hint

Show first that there exists an N 1 such that |an| > |L|/2 for n ≥ N 1.)

Solution: Idea and proof strategy. To show that limn→∞

1

an= 1

L, we need to bound the difference | 1

an− 1

L| by

. We compute:

(0) 1

an−

1

L =

L − ananL

.

The numerator, L − an, is harmless: It can be made smaller than (in absolute value) by our assumption thatlimn→∞

an = L.

The denominator anL in (0), however, could cause problems as an may be close to 0. Thus, we need to ensurethat an cannot get too small. Here is one way to do this: Since {an} converges to L and, by assumption, L = 0from some point onwards, an has to be within |L|/2 of L, and hence must satisfy |an| ≥ |L|/2 (draw a picture tosee this!). With this lower bound the denominator in (0) becomes ≥ |L|2/2 in absolute value, and if we work with|L|2/2 instead of , we obtain as overall bound for the right-hand side of (0).Here is a formal proof:

Proof: Suppose that limn→∞

an = L, L = 0, and an = 0 for all n ∈ N.

Let > 0 be given. Since limn→∞

an = L, applying the definition of a limit with = |L|2/2 (which is > 0 since

L = 0), we obtain an N 1 ∈ N such that

(1) |an − L| < |L|2

2 for all n ≥ N 1.

Also, applying the definition of a limit with = |L|/2, we obtain an N 2 ∈N

such that

(2) |an − L| < |L|

2 for all n ≥ N 2,

If L > 0, then (2) implies an > L − |L/2| = |L|/2 > 0, and if L < 0, (2) implies an < L + |L/2| = −|L| + |L|/2 =−|L|/2, so in either case we have

(3) |an| > |L|

2 for all n ≥ N 2.

4




Now, let N = max(N 1, N 2). Then, for n ≥ N , we have 1

an−

1

L

= |L − an|

|Lan|

< |L|2

2|Lan| (by (1))

< |L|2

2|L|(|L|/2) (by (3)).

=

Hence n ≥ N implies |(1/an) − (1/L)| < .Therefore, lim

n→∞

1/an = 1/L.

Remarks: The second application of the convergence (with |L|/2 playing the role of ) in the above argumentwas necessary to get a lower bound on the denominator anL in (0).

(b) Product property. Prove that if limn→∞


bn = M , then limn→∞

(anbn) = LM . (You can use the result

above that a convergent sequence is bounded.) (Hint: Use a bit of algebraic magic: ab − cd = a(b − d) + d(a − c).)

Solution: Idea and proof strategy. We try to proceed as in the proof of the sum property, but there are twoadditional difficulties we need to overcome.

First, we need to relate the difference anbn− LM (which we seek to estimate) to the differences an− L and bn− M

(which we know how to estimate, by our assumptions that an → L and bn → M ). This can be done by a variationof the familiar trick of adding and subtracting the same term: Namely

anbn − LM = (an − L + L)bn − LM = (an − L)bn + L(bn − M ).

(This a trick well worth remembering!) Applying the triangle inequality as usual, we get a bound involving thedifferences |an − L| and |bn − M |, but now another difficulty appears:

These differences come attached with factors, |bn| and |L|, that we also need to estimate. The second of thesefactors, |L|, is no problem, since it is a constant. The first one requires a bit more thought. One way to deal withthis factor would be to apply the result that any convergent sequence is bounded. An alternative approach thatdoes not depend on this result is to use the add/subtract trick and the triangle inequality in the familiar way|an| = |an − L + L| ≤ |an − L| + |L|. The latter expression can be made ≤ 1 + |L| by applying the definition olimn→∞

an = L with = 1.

Here is the formal proof using the latter approach:Proof: Suppose lim

n→∞


bn = M .

Let > 0 be given, and define by

(1) = min

|L| + |M | + 1, 1

.

Note that this definition ensures that 0 < ≤ 1 and that (|L| + |M | + 1) ≤ .Since lim

n→∞


bn = M , there exist N 1, N 2 ∈ N such that

|an − L| < for all n ≥ N 1,(2)

|bn − M | < for all n ≥ N 2.(3)

From (3) we get, for all n ≥ N 2,

|bn| = |bn − M + M |

≤ |bn − M | + |M | (by triangle inequality)

< + |M | (by (3))

≤ 1 + |M | (by (1)).(4)

Let N = max(N 1, N 2). Then, for n ≥ N we have

|anbn − LM | = |(an − L)bn + L(bn − M )| (by algebra)

≤ |(an − L)bn| + |L(bn − M )| (by triangle inequality)

≤ |bn| + |L| (by (2) and (3))

< (|M | + 1 + |L|) (by (4))

≤ (by (1)).

5




Hence n ≥ N implies |anbn − LM | < .This proves lim

n→∞

(anbn) = LM .

4. Examples and counterexamples. These problems are intended to develop some intuition about the behavior ofsequences. All are quite easy.

(a) Multiplication by bounded sequence. Give an example of a convergent sequence an and a bounded sequence

bn such that the product sequence anbn, does not converge. (The conclusion is true if the sequence an covnergesto 0, but not in general.)

Solution: Example: an = 1, bn = (−1)n. Then an converges with limit 1, bn is bounded, but the productsequence, anbn = (−1)n, does not converge.

(b) Preservation of inequalities. Give an example of sequences an and bn satisfying an < bn for all n ∈ N such thatlimn→∞

an = limn→∞

bn. (We know that an < bn implies (∗) limn→∞

an ≤ limn→∞

bn by an earlier problem. The example

thus shows that the inequality sign in (∗) cannot be replaced by a strict inequality.)

Solution: Example: an = 1 − 1/n, bn = 1 + 1/n. These sequences satisfy an < bn for all n, but they both havethe same limit 1.

(c) Reciprocal property. Give an example of a convergent sequence an satisfying an = 0 for all n for which thereciprocal sequence 1/an is not convergent.

Solution: Example: an = 1/n. This sequence converges with limit 0, but the reciprocal sequence, 1/an = n, doesnot converge.

(d) Sequences with convergent subsequences. Given an example of a sequence an such that the subsequenceover odd-indexed and even-indexed terms, bn = a2n−1 and cn = a2n, both converge, but the sequence an itseldoes not converge.

Solution: Example: an = (−1)n. This sequence does not converge, but the subsequences a2n−1 = −1 and a2n = 1are constant sequences and thus converge.

6

Epsilon Ics 1 Sol

Documents

Transcript of Epsilon Ics 1 Sol