Math 508: Advanced Analysis

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

1 Some limits

(a) Calculate lim n→∞5n+17- n+2

We first see that
5n + 17 5n+ 17 (1∕n) 5+ 17∕n -------= ------- 1-- = -------- n + 2 n+ 2 ∕n 1+ 2∕n

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

5n + 17 -n-+-2-→ 5

(b) Calculate lim n→∞3n2-2n+17- n2+21n+2

Similar to the previous part, we first see
3n2 - 2n+ 17 3- 2∕n + 17∕n2 -n2 +-21n-+-2 = 1+-21∕n-+-2∕n2

Since 1∕n → 0, then 1∕n2 → 0. This implies 3 - 2∕n + 17∕n2 → 3 and 1 + 21∕n + 2∕n2 → 1. Hence 2 3-1+22∕1n∕+n1+72∕∕nn2- converges to three and thus so does 3n2-2n+17- n2+21n+2, as per the opening equation.

2 Calculate lim n→∞√n2--+-n- - n

We first note that
∘ ------ ∘ ------ √n2-+-n+ n n2 + n- n = n2 + n- n √n2-+-n+-n 2 2 = n√--+-n--n-- n2 + n+ n = √----n----- n2 + n+ n ( ) = √----n----- 1∕n- n2 + n+ n 1∕n ----1----- = ∘ ---1- 1+ n + 1
We’ve seen in class that 1n → 0 so that 1 + 1n → 1. Since 1 < 1 + 1n, then 1 < ∘ ----- 1 + 1n < 1 + 1n, which means that ∘ ----1 1 + n → 1 by the “squeeze theorem” in problem 6. It follows that ∘ ----1 1 + n + 1 → 2, from which which see
1 1 ∘--------- → - 1 + 1n + 1 2

Hence, given the opening equation of this problem, we have √ ------ n2 + n - n →12

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 lim n→∞cn n! for c > 0

Fix an integer m so that m > 2c. The for any n > m we have that
c- -c c- 1 1- -1 --1-- m 1- 0 < an = n an- 1 < m an-1 < 2can-1 = 2an-1 < 22an-2 < 23an- 3 < ⋅⋅⋅ < 2n-m am = 2 am 2n

Thus limn→∞an ≤ 2mam limn→∞12n = 0

5

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

(a) Show that lim sup(an + bn) ≤ lim sup an + lim sup n

By definition we have
lim sNu→p∞ (an + bn) = Nlim→∞ (sup (an + bn ) | n > N)

but for any integer N, (sup(an + bn) | n > N ) ≤(sup(an) | n > N ) + (sup(bn) | n > N) so that

lim sup(an + bn) = Nli→m∞ (sup(an + bn) | n > N ) ( ) ≤ Nli→m∞ (sup (an) | n > N)+ (sup(bn) | n > N ) = Nli→m∞ (sup(an) | n > N )+ Nli→m∞ (sup(bn) | n > N ) = lim sup(an) + limsup(bn)

(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
lim ns→up∞(an +bn) = 0

where as

lim sup an + lim sup bn = 1 + 1 = 2 n→∞ n→ ∞

6 Prove a Squeeze Theorem

Let {rn}, {sn}, and {tn} be real sequences such that
rn ≤ sn ≤ tn ∀n
(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 ∈ Br 3(s) for all n ≥ N where N = max(N1,N2). Therefore |rn - tn| < 2r- 3 which, combined with equation 6.1, implies |sn - tn| < 2r 3 for each n ≥ N. Therefore

2r r |sn - s| = |sn - tn + tn - s| ≤ |sn - tn|+ |tn - s| < 3 + 3 = r

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
#(A ∩{1,...,n}) sn := ------n--------

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

sn = 1 - tn
(7.2)

for each n. Now tn = 1 n⌊n⌋ 3, which implies tn ≤1 3 for each n. Thus for any ε > 0 we have

1 1 1 || 1|| ε > 0 = 3 - 3 ≥ tn - 3 = ||tn - 3||

so that tn →1 3. Thus due to Equation 7.2, sn →2 3.

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 = 1 n⌊ n⌋ 2, which with an argument nearly identical to the one for {tn} converging to 1 3 above, we know rn →1 2. 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

a = (1- t )(1 - r ) n n n

Hence an converges to (1 - 13) (1 - 12) = 13 since tn and rn converge to 13 and 12, respectively.

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

Define {xn} recursively by
1 ( A ) xn+1 = 2 xn + x-- n

We note that for the real-valued function f(x) = 1 2(x + A) x we have f′(x) = 1 2(1 - A2) x. Thus f′(x) = 0 when x = -- √ A, implying that f(√A--) = √A-- is the minimum value of f. Since {xn} are just particular values of f, then {xn} is bounded below by √A--.

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

√-- 1 ( A ) √-- 1( √-- A ) √-- 1 ( √-- A ) √ -- y yn+1 = xn+1 - A = - xn + --- - A = - yn + A + -----√-- - A ≤ - yn + A + √--- - A < -n 2 xn 2 yn + A 2 A 2

This then indicates that yn+1 < yn -2 < ⋅⋅⋅ < y1 2n which in turn implies that yn → 0. Since we defined yn = xn -√ -- A, then xn = yn + √ -- A yielding xn →√ -- A.

9

This is not true. We know that the series
∑∞ -1 n=1n

diverges. This implies that

∞∑ --1-- n=1n +1
(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

( an - -1- an ≥ 1 ||{ a + n+11 a ≤ 0 an+1 = {na n++11- a - a > 0 n ||( n n+11 n n- 1 otherwise an - n+1 otherwise

This sequence essentially bounces back and forth between zero and one, so is bounded. It also satisfies the property that |an -an-1| < 1 n since each point an differs by -1- n+1 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 = a1+⋅⋅⋅+an- n, 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
- 1 - 1 - 1,0, 3-,0,-5-,0,...

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
|an - A| < ε 2
(10.4)

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

|an - A| < M
(10.5)

for all n. Thus for n > N we have

a1 +⋅⋅⋅+ aN + aN+1 +⋅⋅⋅+ an cn = -------------n-------------

implying that

|| || |cn - A| = ||a1 +-⋅⋅⋅+-aN-+-aN+1 +-⋅⋅⋅+-an - A || | n | ||a1 +-⋅⋅⋅+-aN-+-aN+1 +-⋅⋅⋅+-an --nA-|| = | n | 1 = n-|(a1 - A )+ ⋅⋅⋅+ (aN - A)+ (aN+1 - A)+ ⋅⋅⋅+ (an - A)| 1 ≤ n-(|a1 - A|+ ⋅⋅⋅+|aN - A|+ |aN+1 - A |+ ⋅⋅⋅+ |an - A|) 1( ε) < n- N M + (n - N)2 1( nε) < n- N M + 2-- N M ε = --n- + 2
by making use of equations 10.4 and 10.5. Thus for any n with n > 2NM- ε we have
|c - A | < N-M-+ ε< NM-ε-+ ε = ε + ε = ϵ n n 2 2NM 2 2 2

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∞1 i. We know that this series diverges, and so the sequence {an} must be unbounded. However, in this case {cn} has
1 ∑n 1 1 ∑n 1 1 cn = -- - ≤ -- --= -- n i=1 i n i=1 n n

which implies cn → 0.