Download MATEM´ATICAS 1215, PARTIAL 3 SOLUTIONS 1. Write down an

MATEMÁTICAS 1215, PARTIAL 3
SOLUTIONS
JOHN GOODRICK
1. Write down an example of a sequence which is bounded (acotada), but
does not converge to any limit.
SOLUTION: One example is the sequence an = (−1)n (i.e. the sequence
−1, 1, −1, 1, . . .). The limit lim an does not exist, but this sequence is bounded
n→∞
above (all the terms are less than 2) and bounded below (all the terms are
greater than −2), so it is bounded.
2. For each of the following series, determine whether it is convergent or
divergent, citing relevat test(s) for convergence.
n
∞ 2
X
n +1
(a)
2n2 − 3
n=1
SOLUTION: We use the Root Test:
s
n
n2 + 1
n2 + 1
n
lim
=
lim
= 1/2.
n→∞
n→∞ 2n2 − 3
2n2 − 3
Since this limit is less than 1, the series converges (absolutely).
(b)
∞
X
(−1)n
n2 + 1
n=1
SOLUTION: This is an alternating series. The function f (x) = x21+1 is
continuous, and limx→∞ x21+1 = 0. To check that f (x) is decreasing on the
interval [1, ∞), note that
2x
f 0 (x) = − 2
,
(x + 1)2
and when x is positive, both 2x and (x2 + 1)2 are positive, so f 0 (x) < 0 when
x > 0. Therefore, by the Alternating Series Test, this series converges.
(c)
∞
X
n=2
n
(ln(n))2
1
2
JOHN GOODRICK
SOLUTION: As n goes to infinity, the terms of the series (by L’Hopital’s
Rule) tend to
n
n
1
n
1
= lim
= lim
= lim = ∞.
= lim
2
n→∞ (ln(n))
n→∞ 2 ln(n)
n→∞ 2 · (1/n)
n→∞ 2
n→∞ 2 ln(n)1/n
Since the limit of the terms is not zero, by the Test for Divergence, this series
diverges.
lim
3. (a) Show that the series
∞
X
2
ne−n converges.
n=1
2
SOLUTION: Let f (x) = xe−x . The function f is continuous, and f (x) > 0
2
if x > 0 (since both x and e−x are positive, and so their product f (x) is
positive). Also, the function f (x) is decreasing on [1, ∞) because
2
2
2
f 0 (x) = e−x − 2x2 e−x = e−x (1 − 2x2 ),
2
and if x ≥ 1 then 2x2 ≥ 2, so 1 − 2x2 < 0, and also e−x > 0 (since e raised to
any power is positive), so their product f 0 (x) is negative.
So by the Integral Test, the series converges if and only if the integral
Z ∞
2
xe−x dx
1
converges. Using the substitution u = x2 , du = 2 dx, the integral is
Z t
Z t2
1 −u
−x2
xe
e du
lim
dx = lim
t→∞ 1
t→∞ 1
2
1 −t2 1 −1
1
= lim − e + e
= .
t→∞
2
2
2e
Since this limit exists (and is finite), the series converges.
(b) Find a number B such that
∞
3
X
X
2
2
−n
−n ne
−
ne ≤ B.
n=1
n=1
The quantity within the absolute value signs is the remainder R3 for the
sum. By the remainder estimate for the Integral Test (see section 11.3 of the
textbook),
Z ∞
Z t
Z t2
1 −u
−x2
−x2
|R3 | ≤
xe
dx = lim
xe
dx = lim
e du
t→∞ 3
t→∞ 9
2
3
1 −t2 1 −9
1
= lim − e + e
= 9.
t→∞
2
2
2e
MATEMÁTICAS 1215, PARTIAL 3
So we can let B =
SOLUTIONS
3
1
.
2e9
∞
X
4n
4. Find the value of the sum
.
5n+1
n=2
SOLUTION: The first few terms of this series are
42 43 44
42
4 42
+
+
+
+
.
.
.
=
(1
+
+ . . .).
53 54 55
53
5 52
The series inside the parentheses is a geometric series, with r = 54 . So the
value of this series is
42
1
16
42
·
·5= .
=
4
3
3
5 1− 5
5
25
5. Let f (x) = ln(1 + 2x) (considered as a function from the real numbers
into the real numbers).
(a) Find the first four terms of the Maclaurin series of f .
SOLUTION: Recall that the nth term of the Maclaurin series is
The first three derivatives of f are:
f 0 (x) =
f 00 (x) = −
f (n) (0) n
x .
n!
2
= 2(1 + 2x)−1
1 + 2x
4
= −4(1 + 2x)−2
2
(1 + 2x)
16
= 16(1 + 2x)−3
(1 + 2x)3
So the first four terms of the Maclaurin series are
f (3) (x) =
f (0) + f 0 (0)x +
f 00 (0) 2 f (3) (0) 3
8
x +
x = 0 + 2x − 2x2 + x3 .
2
6
3
(b) Find the radius of convergence of the Maclaurin series of f .
SOLUTION: If n ≥ 1, then continuing the pattern from part (a), we see
that the nth derivative of f is
f (n) (x) = (−1)n (n − 1)!2n (1 + 2x)−n .
So the nth term of the Maclaurin series is
2n
f (n) (0) n
x = (−1)n xn .
n!
n
4
JOHN GOODRICK
We use the Ratio Test to find values of x for which the Maclaurin series
converges:
n+1
n
2
n+1
2n|x|
n+1
n
n+1
x
· (−1) n n = lim
= 2|x| lim
= 2|x|.
lim (−1)
n→∞
n→∞
n→∞ n + 1
n+1
2 x
n
Since the Maclaurin series converges when this limit is less than 1 and diverges when it is greater than 1, we see that the series converges for x in the
interval (−1/2, 1/2) and diverges when x < −1/2 or when x > 1/2. Therefore
the radius of convergence is 1/2.
6. Find all values of x for which the following series converges:
∞
X
3n xn
√
.
3
n
n=1
(HINT: It is not enough to find the radius of convergence. You must also test
whether the series converges at the endpoints of the radius of convergence.)
SOLUTION: First we find the radius of convergence using the Ratio Test:
n+1 n+1 √
√
3
3
3 x
n
3|x|
n
= lim √
lim √
·
3
3
n
n
n→∞
n→∞
n+1 3 x
n+1
r
r
n
n
= 3|x| 3 lim
= 3|x|.
= 3|x| lim 3
n→∞
n→∞ n + 1
n+1
Therefore the radius of convergence is 1/3.
Next, we need to check whether the series converges when x = ±1/3. If
x = −1/3, then the series is alternating, and
3n (−1/3)n
(−1)n
√
√
lim
=
lim
= 0.
3
3
n→∞
n→∞
n
n
1
Also, the function f (x) = √
3 x is continuous and decreasing on [1, ∞), so by the
Alternating Series Test, this series converges.
If x = 1/3, then the series becomes
∞
X
1
√
.
3
n
n=1
∞
X
1
If n ≥ 1, then n ≤ n, so
≥
Since the harmonic series
diverges,
n
n=1
by the Comparison Test, this series diverges, too.
In conclusion, the series converges when x is in the interval [−1/3, 1/3) and
diverges otherwise.
√
3
1
√
3n
1
.
n