18.014 Problem Set 11 Solutions
Sam Elder
December 7, 2015
Problem 1. Let f : R → R be defined by
(
f (x) =
e−1/x for
0 for
x>0
x ≤ 0.
Part 1.1. Prove that f is continuous at 0.
Solution. We must show that lim f (x) = 0. Clearly lim− f (x) = lim− 0 = 0, so it suffices to show that
x→0
x→0
x→0
lim f (x) = lim e−1/x = 0. Replacing x with 1/x, this is equivalent to lim e−x = 0, which we already
x→0+
x→∞
x→0+
know from class (or equation (7.30) in the book).
Part 1.2. Prove that f is differentiable at 0.
f (x)
exists, and again we will split into the left and right limits Since
x
0
e−1/x
= lim− = 0. Indeed, replacing x with
f (x) = 0 for x ≤ 0, this amounts to showing that lim+
x
x→0 x
x→0
x
1/x again, this is equivalent to showing that lim x = 0, which follows from Theorem 7.11 in the book (or
x→∞ e
L’Hôpital’s Rule).
Solution. We must show that lim
x→0
Part 1.3. Determine whether or not the derivative of f is continuous at 0.
Solution. We claim that it is. First, we compute the derivative at x 6= 0, since we have shown that f 0 (0) = 0
e−1/x
. To show
in the previous part. Clearly f 0 (x) = 0 for x < 0. By the chain rule, for x > 0, f 0 (x) =
x2
0
0
that lim f (x) = f (0) = 0, we again split into the left and right limits. The left limit is trivially zero, so we
x→0
must show that
lim+
x→0
e−1/x
= 0.
x2
x2
We again replace x with 1/x. Then the limit is equivalent to lim x = 0, which also follows from Theorem
x→∞ e
7.11 or L’Hôpital’s Rule twice.
Problem 2 (10.4-7). Determine whether this sequence converges or diverges, and find the limit if it con1 + (−1)n
verges: f (n) =
.
n
Answer. This converges to 0.
1+1
2
Solution. Notice that |f (n)| ≤
=
by the chain rule. Therefore, for > 0, let N = d2/e. If
n
n
2
2
n > N ≥ 2/, then |f (n)| ≤ <
= , so f (n) → 0 as desired.
n
2/
1
Problem 3 (10.9-3). Prove that
∞
X
n=2
n2
1
3
= .
−1
4
Solution 1. Combining pairs of consecutive terms. The partial fraction decomposition of this ratio is
1
1
1
1
=
−
.
n2 − 1
2 n−1 n+1
This series isn’t quite the form of telescoping we’ve seen, but we can instead pair up consecutive terms in
the sum:
∞
∞ X
X
1
1
1
=
+
n2 − 1
(2k)2 − 1 (2k + 1)2 − 1
n=2
k=1
∞ 1X
1
1
1
1
=
+
−
−
2
2k − 1 2k 2k + 1 2k + 2
k=1
1
1
1
1
=
1 + − lim
+
2
2 k→∞ 2k − 1 2k
1
3
1
=
1+ −0 = ,
2
2
4
P∞
1
1 1
by telescoping, as the sum is equal to k=1 (ak − ak+1 ) for ak = 2k−1
+ 2k
.
Solution 2. Some clever rearranging provides a slightly more rigorous solution. We again start with the
partial fraction decomposition:
1
1
1
1
1
1
1
1 1
1
=
−
=
+
−
+
.
n2 − 1
2 n−1 n+1
2 n−1 n
2 n n+1
1
1
1
+
In this way, we have written the terms of this series as an − an+1 , where an =
. Therefore,
2 n−1 n
∞
X
∞
X
1
1
1
1
3
=
(an − an+1 ) = a2 − lim an =
+
−0= .
2
n→∞
n − 1 n=2
2 2−1 2
4
n=2
Problem 4. Compute the sum of the series
∞
X
π 4n
.
(4n)!
n=0
1 This rewriting technique does not always work for infinite series; it would also claim that 1 − 1 + 1 − 1 + 1 − . . . would
converge, which it does not. However, an easy argument involving the sequence of partial sums shows that it works when all
terms of the series are positive, or if the sequence itself converges to zero, both of which are satisfied here.
2
Solution. We start with the series for ex , e−x , and cos x:
∞
X
x2
x3
x4
xn
e =
=1+x+
+
+
+ ···
n!
2!
3!
4!
n=0
x
e−x =
cos(x) =
cosh(x) =
∞
X
x2
x3
x4
(−x)n
=1−x+
−
+
− ···
n!
2!
3!
4!
n=0
∞
X
(−1)n x2n
x2
x4
=1−
+
− ···
(2n)!
2!
4!
n=0
∞
X
1 x
x2
x4
x2n
(e + e−x ) =
=1+
+
+ ···
2
(2n)!
2!
4!
n=0
∞
X
1
x4
x8
x4n
(cosh(x) + cos(x)) =
=1+
+
+ ··· ,
2
(4n)!
4!
8!
n=0
so the desired sum is 12 (cosh π + cos π) =
eπ + e−π
1
− .
4
2
Problem 5. Define a sequence {an } by a1 = 2 and
an+1 =
an + 2/an
2
√
for n ∈ N. The goal of this problem is to prove that this sequence converges to 2.
√
√
Part 5.1. Prove that an ≥ 2 for all n by showing that x + x2 ≥ 2 2 for any positive real x.
√
Solution. This clearly holds for x > 2 2 and x < √12 , since one of those two positive terms will individually
√ √
minimum must occur at either an
point.
be greater. On x ∈ [1/ 2, 2 2], the √
√ endpoint or a critical
√
√ Since
√
(x + 2/x)0 = 1 − 2/x2 = 0 when x = 2, the minimum occurs at x = 2, in which case 2 + 2/ 2 = 2 2
as desired.
Part 5.2. Use the previous part to show that the sequence {an } is decreasing.
Solution. Since an ≥
decreasing.
√
2, 2/an ≤
√
2 ≤ an , so an+1 =
an + an
an + 2/an
≤
= an and the function is indeed
2
2
√
Part 5.3. Conclude that the sequence must converge and prove that it converges to 2.
√
Solution 1. The sequence is monotone decreasing, but bounded below by 2, so it must have a limit L ≥
√
2 > 0. Therefore,
1
2
1
2
lim an+1 =
lim an +
=
lim L +
.
n→∞
2 n→∞
an
2 n→∞
L
But limn→∞ an+1 = limn→∞ an = L as well, so we have
√
1
2
2
L+
=⇒ L =
=⇒ L = 2,
L=
2
L
L
since we know that L > 0, as desired.
3
Solution 2. Another approach offers a crude bound on how quickly it converges. We write it this way:
√
√
√
√
an − 2 2 + 2/an
a2 − 2 2an + 2
(an − 2)2
an+1 − 2 =
= n
=
.
2
2an
2an
√
√
√
Since an ≥ 2 > 1 and an − 2 ≤ 2 − 2 < 1, this implies that
√
√
an − 2
an+1 − 2 <
,
2
√
√
a1 − 2
1
< n−1 . Since this fraction goes to 0 as
from which an easy induction implies that an − 2 <
n−1
2√
2
√
n → ∞ and an > 2 for all n, we have proven that an → 2, as desired.
4