HOMEWORK 1 SOLUTIONS
Problem 1
We want to show that multiplication in the reals is associative. This amounts
to showing that for any given x, y, z ∈ R the following limits coincide.
lim tn (xy)tn (z)
n→∞
and
lim tn (x)tn (yz)
n→∞
to do this we will compare both tn (xy)tn (z) and tn (x)tn (yz) to the quantity
tn (x)tn (y)tn (z). Note that we do not have to worry about inserting parenthesis in
the last quantity because tn (x), tn (y) and tn (z) are all rational numbers and we
know that the rationals satisfy associativity of multiplication. Now suppose that
we have two real numbers a and b. We want to be able to compare tn (ab) with
tn (a)tn (b). Note that ab can be written as
ab = (tn (a) + (a − tn (a)))(tn (b) + (b − tn (b)))
= tn (a)tn (b) + tn (a)(b − tn (b)) + tn (b)(a − tn (a)) + (a − tn (a))(b − tn (b))(∗)
Now note that since |b − tn (b)| ≤ 1/10n and similarly |a − tn (a)| ≤ 1/10n we get
that (∗) implies
|ab − tn (a)tn (b)| ≤ |tn (a) + tn (b)|/10n + 1/102 n < (|tn (a)| + |tn (b)| + 1)/10n
Also note that |ab−tn (ab)| ≤ 1/10n and so putting these together using the triangle
inequality gives
|tn (ab)−tn (a)tn (b)| = |tn (ab)−ab+ab−tn (a)tn (b)| ≤ |tn (ab)−ab|+|ab−tn (a)tn (b)|
≤ (|tn (a)|+|tn (b)|+1)/10n +1/10n = (|tn (a)|+|tn (b)|+2)/10n < (|a|+|b|+2)/10n
. Using the above inequality we can solve the problem. For any x, y and z in R we
get
|tn (x)tn (yz) − tn (x)tn (y)tn (z)| = |tn (x)||tn (yz) − tn (y)tn (z)|
≤ |tn (x)|(|y| + |z| + 2)/10n ≤ (|xy| + |xz| + 2|x|)/10n
and similarly
tn (xy)tn (z) − tn (x)tn (y)tn (z)| ≤ (|zx| + |zy| + 2|z|)/10n
putting these two together using the triangle inequality now gives
|tn (x)tn (yz) − tn (xy)tn (z)| ≤ (|xy| + |zy| + 2(|xz| + |x| + |z|))/10n
So we have shown that |tn (x)tn (yz) − tn (xy)tn (z)| < C/10n where C = (|xy| +
|zy| + 2(|xz| + |x| + |z|)) is a constant. It now will follow that tn (x)tn (yz) and
tn (xy)tn (z) must converge to the same limit. To see this note that x(yz) is the
limit of the sequence {tn (x)tn (yz)} and we want to show that it is also the limit of
the sequence {tn (xy)tn (z)}.
Date: October 19, 2013.
1
2
HOMEWORK 1 SOLUTIONS
Suppose we are given ε > 0 then choose N large enough so that for n > N we
have |x(yz) − tn (x)tn (yz)| < ε/2 and also C/10N < ε/2. Then for n > N we get
|x(yz) − tn (xy)tn (z)| ≤ |x(yz) − tn (x)tn (yz)| + |tn (x)tn (yz) − tn (xy)tn (z)|
< ε/2 + C/10n < ε/2 + ε/2 = ε
So we have shown that the limit of the sequence {tn (xy)tn (z)} is precisely x(yz)
and we are done.
Problem 2
Since {an } and {bn } are Cauchy sequences, they must converge. So, lim an
n→∞
and lim bn exists (i.e. they are real numbers, and in particular, finite). Let
n→∞
lim an = L1 and lim bn = L2 .
n→∞
n→∞
[Now the idea is that, by definition of limit an ’s go arbitrarily close to L1 and
bn ’s go arbitrarily close to L2 , while by the given condition an ’s go arbitrarily close
to bn ’s and therfore L1 must go arbitrarily close to L2 . If two real numbers are
abitrarily close, they must be equal (why?). So we need something that tells us, if
x is close to y, y is close to z and z is close to w then x is close to w. What very
powerful (but deceptively simple) theorem tells us things like this? The triangle
inequality. It says if x is close to y and y is close to z then x is close to z. So, we
just need to apply the triangle inequality twice.]
So, let’s make this formal.
Take any > 0.
By definitiion of limit, ∃N1 , N2 ∈ N such that ∀n > N1 , m > N2 we have
|an − L1 | < and |bm − L2 | < .
3
3
3
Now, let N ∈ N be such that N > . (Why must such a N always exist?)
1
So,
<
N
3
1
Now, by the given condition, |aN − bN | ≤
< .
N
3
1
1
Also, by the given condition, ∀j > N, |aj − bj | ≤ <
< .
j
N
3
Let k = 1 + max{N, N1 , N2 }.
Therefore, |ak − L1 | <
, |bk − L2 | < , |ak − bk | < . (why?).
3
3
3
[Do you see why I chose k = 1+max{N, N1 , N2 } instead of k = max{N, N1 , N2 }?]
Now, using the triangle inequality twice,
|L1 − L2 | ≤ |L1 − ak | + |ak − L2 | ≤ |L1 − ak | + |ak − bk | + |bk − L2 | <
+ + =
3 3 3
HOMEWORK 1 SOLUTIONS
3
[Do you now see why I have been putting a 3 under the everywhere?]
But is an arbitrary positive real number. Therefore, we can conclude that,
|L1 − L2 | is smaller than all positive real numbers (why?). But |L1 − L2 | is nonnegative by definition of absolute value. A non-negative real number can be smaller
than all positive real numbers only if it is 0 (why?).
Therefore, |L1 − L2 | = 0 =⇒ L1 − L2 = 0 =⇒ L1 = L2 .
Therefore, the two sequences {an } and {bn } converges to the same limit.
1. Problem 3
Note that some digit d1 must occur infinitely often in the first decimal place
of the terms in the sequence {cn }. We define a subsequence {ci1,k } of {cn } by
taking the terms with the first decimal place equal to d1 . Having defined {cij,k },
let us define a subsequence {cij+1,k } of {cij,k } as follows: some digit dj+1 must
occur infinitely often in the j + 1 decimal place of the terms in the sequence {cij,k },
we take {cij+1,k } to be the subsequence consisting of the terms in {cij,k } with the
(j + 1)-th decimal place equal to dj+1 .
Now, we define a subsequence {cik } of the original sequence {cn } by taking
cik = cik,k . We shall show that this subsequence converges to d := 0.d1 d2 d3 . . .. For
any > 0, we can find N so that 10−N < . Then if k > N , we have
|cik − d| = |cik,k − d| < 10−k < 10−N < ,
so limk→∞ cik = d ∈ C, as desired.
Problem 4
∞
X
i=2
i
i+1
∞ 2X
−1
X
1
1
=
k
n(log2 n)k
n(log
2 n)
i
i=1
n=2
i+1
where k = 1, 2. For 2 ≤ n < 2 , i ≤ log2 n < i + 1.
When k = 1, we want to show this series diverges, we want to bound it below
by some divergent series.
∞
X
i=2
∞
P
i=2
i+1
i+1
n=2
n=2
∞ 2X
−1
∞ 2X
−1
∞
∞
X
X
X
X
1
2i
1
1
1
=
>
=
=
i+1 (i + 1)
i+1 (i + 1)
n log2 n
n
log
n
2
2
2(i
+ 1)
2
i
i
i=1
i=1
i=1
i=1
1
n log2 n
is bounded below by 1/2 of harmonic series without the first term which
diverges, so it diverges itself.
When k = 2, we want to show this series converges, we want to bound it above
by some convergent series.
∞
X
i=2
∞
P
i=2
i+1
i+1
n=2
n=2
∞ 2X
−1
∞ 2X
−1
∞
∞
X
X
X
X
1
1
1
2i
1
π2
=
≤
=
=
=
2
2
i
2
i
2
2
n(log2 n)
n(log2 n)
2i
2i
i
6
i
i
i=1
i=1
i=1
i=1
1
n(log2 n)2
is increasing and bounded above by a constant, thus it converges.
4
HOMEWORK 1 SOLUTIONS
Problem 5
Let
Sn =
n2013 2013n
n!
then we have
(n + 1)2013 2013.2013n n2013 2013n
/
(n + 1)n!
n!
2013
n+1
2013
=
n
n+1
Sn+1 /Sn =
Now using the fact that limn→∞ (xn .yn ) = limn→∞ (xn ) limn→∞ (yn ) for two convergent sequences we get:
2013
2013
n+1
n+1
lim (
) = lim (
)
=1
n→∞
n→∞
n
n
Sn+1
2013
lim (
)=0
) = 1. lim (
n→∞ Sn
n→∞ n + 1
By ratio test the series converges.