Homework 2, Math 4111, due September 12
P
1
(1) Let xn = nk=1 k(k+1)
. Prove that lim xn = 1. (Hint:
1
1
− k+1 .)
k
xn =
n
X
k=1
n
X
1
=
k(k + 1) k=1
1
1
−
k k+1
=1−
1
k(k+1)
=
1
.
n+1
Rest is easy.
P
(2) Let xn = nk=1 k12 . Prove that lim xn exists. (Hint: The set
1
{xn } is bounded above since k12 ≤ k(k−1)
for k ≥ 2 and the
previous problem.)
This is easy. Using the hint, we see that |xn | ≤ 2 for all
n. This is a monotonic increasing sequence and thus by the
theorem proved in class, lim xn is just the supremum of the set
{xn }.
(3) Assume lim xn = a, lim yn = b. Prove that lim xn + yn = a + b
and lim xn yn = ab. (Hint: xn yn − ab = (xn − a)yn + a(yn − b)
and both {xn }, {yn } are bounded.)
Let > 0 be given. Then we can find N1 , N2 ∈ N so that for
all n ≥ N1 , |xn − a| < /2 and for all m ≥ N2 , |ym − b| < /2.
So, if we set N = max(N1 , N2 ), then for all n ≥ N , we have,
|xn + yn − (a + b)| = |(xn − a) + (yn − b)|
≤ |xn − a| + |yn − b| < /2 + /2 = Again, let > 0 be given and let M ∈ R be some positive
number which is greater than an upper bound for |yn | for all n
(which exists, since {yn } is a CS) and |a|. Then, we can find
N1 , N2 ∈ N so that for all n ≥ N1 , |xn − a| < /2M and for all
≥ N2 , |yn − b| < /2M . St N = max(N1 , N2 ). Then for any
n ≥ N,
|xn yn − ab| = |(xn − a)yn + a(yn − b)|
≤ |yn ||xn − a| + |a||yn − b|
< M /2M + M /2M = Pn 1
(4) Let xn =
k=1 k . Prove that the set {xn } is not bounded
P m+1
above. (Hint: 2k=2m +1 k1 ≥ 12 .)
Let am be the sum in the hint. If 2m + 1 ≤ k ≤ 2m+1 , then
1
1
≥ 2m+1
and thus am is at least 1/2, since all the terms are at
k
1
2
1
2m+1
and there are precisely 2m terms. For any m ∈ N,
m
x2m = 1 + a0 + a1 + · · · + am−1 ≥ 1 + .
2
Since m ∈ N is unbounded, so are the xn ’s.
Pn
xk
(5) Let lim xn = a and let {yn } be the sequence, yn = k=1
.
n
Prove that lim yn = a.
ease of notation, let bn = xn − a. Then yn − a =
PFor
n
k=1 bk /n. Let > 0 be given. Then we can find an N1 ∈ N so
that for all n ≥ N1 , |bn | = |xn − a| < /2. We can also choose
an N2 ∈ N so that
b1 + b2 + · · · + bN 1 < /2.
(1)
N2
Notice that N1 is fixed and so we can choose N2 to be large
enough to achieve this. Let N = max(N1 , N2 ). Then for any
n ≥ N,
least
Pn
k=1 bk |yn − a| = n
P
Pn
N1 b
k=1 k
k=N1 +1 bk =
+
n
n
Pn
P
N1 b |bk |
k
≤ k=1 + k=N1 +1
n
n
The
P first sum on the right hand side is less than or equal
N1 b k since n ≥ N ≥ N2 and thus by equation (1), this
to k=1
N2
is less than /2. On the other hand, in the second sum on
the right, since k > N1 for the bk ’s occurring in that sum, we
have |bk | < /2 and therefore, the sum is less than or equal to
(n−N1 )
< /2. Thus, we get that,
2n
|yn − a| < /2 + /2 = ,
for all n ≥ N .