Homework 4 <me@tylerlogic.com>
September 29, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework04

### 1 Some limits

#### (a) Calculate limn→∞

We first see that

Since 1∕n 0 then 17∕n 0 and 2∕n 0. Therefore 5 + 17∕n 5 and 1 + 2∕n 1. Thus 1 implying that 5. Given the opening equation, this then final implies

#### (b) Calculate limn→∞

Similar to the previous part, we first see

Since 1∕n 0, then 1∕n2 0. This implies 3 - 2∕n + 17∕n2 3 and 1 + 21∕n + 2∕n2 1. Hence converges to three and thus so does , as per the opening equation.

### 2 Calculate limn→∞- n

We first note that
We’ve seen in class that 0 so that 1 + 1. Since 1 < 1 + , then 1 < < 1 + , which means that 1 by the “squeeze theorem” in problem 6. It follows that + 1 2, from which which see

Hence, given the opening equation of this problem, we have - n

### 3

Let {an > 0} be a sequence of reals converging to a real A > 0. Then BA∕2(A) contains all but finitely many points of {an}. Let a be the minimum value of the points not in BA∕2(A). Note this point is positive since all an are positive. Thus setting c = min(a,A∕2)2 we have that an > c for all n.

### 4 Show that limn→∞ for c > 0

Fix an integer m so that m > 2c. The for any n > m we have that

Thus limn→∞an 2mam limn→∞ = 0

### 5

Let {an} and {bn} be real sequences.

#### (a) Show that limsup(an+ bn) ≤limsupan+limsupn

By definition we have

but for any integer N, + so that

#### (b) Give an explicit example where strict inequality can occur.

Let {an} be (-1)n and {bn} be (-1)n+1 so that {an} has value 1,-1 on even, odd indices, respectively, but {bn} has value 1,-1 on odd, even indices, respectively. With these sequences

where as

### 6 Prove a Squeeze Theorem

Let {rn}, {sn}, and {tn} be real sequences such that
 (6.1)

Assume that both tn s and rn s. Then for any r > 0, there is are integers N1,N2 such that rn,tn B(s) for all n N where N = max(N1,N2). Therefore |rn - tn| < which, combined with equation 6.1, implies |sn - tn| < for each n N. Therefore

for each n N. Hence Br(s) contains all but finitely many points in {sn}, so that sn s.

### 7

Let A be the set of natural numbers not divisible by three. Define

be the fraction of natural numbers from one to n which are not divisible by three. Thus defining the sequence {tn} where tn is the fraction of natural numbers less than or equal to n that are divisibile by three, we have

 (7.2)

for each n. Now tn = , which implies tn for each n. Thus for any ε > 0 we have

so that tn . Thus due to Equation 7.2, sn .

Repeat this for all natural numbers not multiples of two or three Let {tn} again be as above, the fraction of natural numbers less than or equal to n which are divisible by three. Define {rn} as the sequence where each rn is the fraction of natural numbers less than or equal to n which are divisible by two. Thus we have rn = , which with an argument nearly identical to the one for {tn} converging to above, we know rn . If, now, the sequence {an} is such that each an is the fraction of natural numbers less than or equal to n which are not divisible by two or three, we have

Hence an converges to = since tn and rn converge to and , respectively.

### 8 Show the sequence of terms in Newton’s method for solving x2- A = 0 converges to

Define {xn} recursively by

We note that for the real-valued function f(x) = we have f(x) = . Thus f(x) = 0 when x = , implying that f() = is the minimum value of f. Since {xn} are just particular values of f, then {xn} is bounded below by .

So let’s define a sequence {yn} by yn = xn -. Then we have

This then indicates that yn+1 < < < which in turn implies that yn 0. Since we defined yn = xn -, then xn = yn + yielding xn .

### 9

This is not true. We know that the series

diverges. This implies that

 (9.3)

also diverges since it’s one less than the first. We use this fact to our advantage in finding a counter-example. Define the sequence {an} by recursively by a0 = 0 and

This sequence essentially bounces back and forth between zero and one, so is bounded. It also satisfies the property that |an -an-1| < since each point an differs by from the previous. Furthermore, we know once the sequence “hits” one and heads back to zero (and vice versa), that it will actually reach zero because of the divergence of 9.3 above; i.e. starting at zero but not being able to reach one (and vice versa) at any point in the sequence would imply convergence of 9.3.

### 10

For a real sequence {an} define {cn} by cn = , i.e. {cn} is the sequence of partial averages of {an}.

#### (a) Find an example where {an} doesn’t converge, but {cn} does.

Let an = (-1)n. Then {an} doesn’t converge, bouncing back and forth between 1 and -1, but {cn} is the sequence

converging to zero.

#### (b) Show If cn→ A whenever an→ A

Let an A. For ε > 0 we can find an integer N such that
 (10.4)

Furthermore, because {an} converges, it is bounded, so that we can find an integer M with

 (10.5)

for all n. Thus for n > N we have

implying that

by making use of equations 10.4 and 10.5. Thus for any n with n > we have

implying the convergence of {cn} to A.

#### (c) If ak≥ 0 and the averages converge, must {ak} be bounded?

No. For example, let {an} be the sequence of partial sums of the harmonic series i=1. We know that this series diverges, and so the sequence {an} must be unbounded. However, in this case {cn} has

which implies cn 0.