Math 101—Practice Final Examination
April 2016
Duration: 150 minutes
Section #: 211
Name: Martin, Greg
Student #: 12345678
(sort #: 4000)
Signature
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Problem Out of Score Problem Out of Score
1
8
7
7
2
8
8
7
3
6
9
7
4
6
10
7
5
6
11
7
6
6
Total
75
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 2 of 12
Problems 1–2 are answer-only questions: the correct answers in the given boxes earns full credit,
and no partial credit is given.
R1
1a. [2 pts] For the integral 0 e−x dx, which of the following approximate integration methods
will give an underestimate for any value of n? Put the letter(s) corresponding to the correct
answer(s) into the box; there can be any number of correct answers. Answer:
L:
R:
S:
T:
R
Riemann sum using left endpoints of subintervals
Riemann sum using right endpoints of subintervals
Simpson’s Rule
Trapezoid Rule
R1
Since e−x is decreasing, a right Riemann sum for 0 e−x dx is an underestimate; the left
Riemann sum would be an overestimate. Since e−x is concave up, a Trapezoid Rule
Riemann sum is an overestimate (and it turns out that a Midpoint Rule Riemann sum
would be an underestimate in this case). There is no geometric way to judge the inequality
between a Simpson’s Rule estimate of an integral and the actual value of the integral.
1√
1b. [2 pts] Which expression is a Riemann sum for
1 − x2 dx? Answer:
C
0
s
2
n
X
i
i
A:
1−
n
n
i=1
s
2
n
X
i
2
1−
B:
n
2n
i=1
s
2
n
X
i−1
1
1−
C:
n
n
i=1
s
2
n
X
i
i
D:
1−
2
n
n
i=1
Z 1√
Sum C is a left Riemann sum for
1 − x2 dx with n subdivisions. Sum B is a right
0
Z 1 p
Z 2p
2
Riemann sum for
1 − (x/4) dx, or for
2 1 − (x/2)2 dx. Sum D is a right
0
0
Z 1 √
Riemann sum for
x 1 − x2 dx. Sum A isn’t a Riemann sum at all.
Z
0
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 2 of 12
1c. [4 pts] Suppose f (x) and g(x) are twice differentiable functions with the following values:
f (0) = −1
f (1) = 3
f (2) = 4
f (3) = 9
f 0 (1) = 2
g(0) = 3
g(1) = 0
g(2) = 1
g(3) = 6
g 0 (1) = −1
1
Z
f 0 (g(x))g 0 (x) dx.
First, evaluate
Answer:
−10
0
Z
Then, evaluate
1
xg 00 (x) dx.
Answer:
2
0
Simplify both answers completely.
Making the substitution u = g(x) in the first integral gives
Z 1
Z g(1)
0
0
f (g(x))g (x) dx =
f 0 (u) du = f (g(1)) − f (g(0)) = −1 − 9 = −10.
0
g(0)
Alternatively, one can simply recognize from the chain rule that f (g(x)) is an antiderivative of f 0 (g(x)g 0 (x), and use the Fundamental Theorem of Calculus immediately to get
the same answer.
For the second integral, integration by parts with u = x and dv = g 00 (x) dx, so that
du = dx and v = g 0 (x), gives
Z 1
Z 1
1
00
0
xg (x) dx = xg (x)0 −
g 0 (x) dx = g 0 (1) − (g(1) − g(0)) = 2.
0
0
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 3 of 12
Problems 1–2 are answer-only questions: the correct answers in the given boxes earns full credit,
and no partial credit is given.
2a. [4 pts] For each of the following series, choose the appropriate statement.
CA: Converges absolutely
CC: Converges conditionally
D: Diverges
∞
X
(−1)n
(i)
Answer:
CC
2n
+
7
n=1
∞
X
(ii)
(−1)n
Answer:
D
n=1
∞
X
(−1)n
(iii)
n3 + 2
n=1
∞
X
cos n
(iv)
2n
n=1
Answer:
Answer:
CA
CA
∞
∞ X
X
(−1)n (−1)n
(i) The series
converges by the Alternating Series Test. But
2n + 7 =
2n
+
7
n=1
n=1
∞
∞
X
X
1
1
diverges, by the Limit Comparison Test with the harmonic series
.
2n
+
7
n
n=1
n=1
∞
X
(−1)n
Therefore,
converges conditionally.
2n
+
7
n=1
∞
X
(−1)n diverges by the Test for Divergence, as lim (−1)n does not
(ii) The series
n=1
equal 0.
n→∞
∞ ∞
X
(−1)n X
1
(iii) The series
converges, by the Comparison Test with the
n3 + 2 =
3
n +2
n=1
n=1
∞
∞
X
X
1
(−1)n
convergent p-series
.
Therefore,
converges absolutely.
n3
n3 + 2
n=1
n=1
∞
X
cos n 1
1
is a convergent geometric series.
(iv) Note that 2n ≤ 2n for every n. The series
2n
n=1
∞ X
cos n Hence by the Comparison Test, the series
n also converges, and therefore
2
n=1
∞
X
cos n
the series
converges absolutely.
n
2
n=1
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 3 of 12
3
2b. [2 pts] Which of the following Maclaurin series equals
x
?
+1
x2
H
Answer:
F : x 2 − x4 + x6 − x8 + · · ·
K : x2 + x4 + x6 + x8 + · · ·
G : x2 − 21 x4 + 14 x6 − 81 x8 + · · ·
M : x2 + 12 x4 + 14 x6 + 18 x8 + · · ·
H : x 3 − x5 + x7 − x9 + · · ·
P : x3 + x5 + x7 + x9 + · · ·
J : x3 − 21 x5 + 14 x7 − 81 x9 + · · ·
Q : x3 + 12 x5 + 14 x7 + 18 x9 + · · ·
∞
X
1
Recall that
=
xn for all |x| < 1. Substituting −x2 in for x, we find that
1 − x n=0
∞
∞
X
X
1
2 n
=
(−x
)
=
(−1)n x2n .
1 + x2
n=0
n=0
Multiplying by x3 on both sides, we conclude that
∞
X
x3
=
(−1)n x2n+3 = x3 − x5 + x7 − x9 + · · · .
1 + x2
n=0
2c. [2 pts] The power series
∞
X
An (x + 2)n converges at x = −4 and diverges at x = 1. What
n=0
are the possible values of its radius of convergence R? Write your answer either in interval
notation, or in the form a ≤ R ≤ b for some numbers a and b.
The power series
∞
X
Answer:
[2, 3]
an (x + 2)n converges if |x + 2| < R; as the series diverges at x = 1,
n=0
we must have that |1 + 2| ≥ R, that is, 3 ≥ R. Also, the power series
∞
X
an (x + 2)n
n=0
diverges if |x+2| > R; as the series converges at x = −4, we must have that |−4+2| ≤ R,
that is, 2 ≤ R. In summary, we’ve found that 2 ≤ R ≤ 3.
(As an exercise, you can check that all of these values of R really are possible, for various choices of the coefficients An . For example, you can choose An = 1/nRn for any
particular 2 ≤ R ≤ 3.)
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 4 of 12
Problems 3–6 are short-answer questions: put a box around your final answer, but you must include
the correct accompanying work to receive full credit.
Z
3a. [3 pts] Find sin2 x cos3 x dx.
As the power of cosine in the integrand is odd, we let u = sin x, so that du = cos x dx
and
Z
Z
2
3
sin x cos x dx = sin2 x · (1 − sin2 x) cos x dx
Z
= u2 (1 − u2 ) du
Z
= (u2 − u4 ) du
u3 u5
sin3 x sin5 x
=
−
+C =
−
+ C.
3
5
3
5
3b. [3 pts] Find the area enclosed by the graphs of y = x2 and y =
completely.
√
x. Simplify your answer
√
These graphs intersect where x2 = x, so that x4 = x; so either x√= 0 or else x3 = 1,
which means x = 1. Between these two intersection points we have x ≥ x2 . Therefore,
the area enclosed is
3/2
1
Z 1
Z 1
√
x
x3 2 1
1
2
1/2
2
( x − x ) dx =
−
= − = .
(x − x ) dx =
3/2
3 0 3 3
3
0
0
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 5 of 12
Problems 3–6 are short-answer questions: put a box around your final answer, but you must include
the correct accompanying work to receive full credit.
x3 + log x
4a. [3 pts] Find the average value of the function
for 1 ≤ x ≤ 3.
x
We know that the average value in question is
Z 3
Z 3
Z 3 3
log x
1
x + log x
1
2
dx =
x dx +
dx .
3−1 1
x
2
x
1
1
R3
3 3
The first integral is easy, as 1 x2 dx = x3 1 = 27
− 13 = 26
. In the second integral, we
3
3
1
let u = log x, so that du = x dx and
log 3
Z 3
Z log 3
log x
u2 (log 3)2
dx =
u du = .
=
x
2 0
2
1
0
(Notice that the endpoints of integration must change with the substitution: when x = 1
we have u = log 1 = 0, while when x = 4 we have u = log 3.) The average value in
question is therefore
Z 3
Z 3
1
log x
1 26 (log 3)2
13 (log 3)2
2
x dx +
dx =
+
+
=
2
x
2 3
2
3
4
1
1
4b. [3 pts] Find the solution of the differential equation
2
xex + 2
y =
y
with initial condition y(0) = −2. Write your answer in the form y = g(x) for some
function g(x).
0
We separate variables:
2
xex + 2
dy
=
dx
y
2
y dy = (xex + 2) dx
Z
Z
2
y dy = (xex + 2) dx.
By using the substitution u = x2 in the integral on the right side, we get
2
y2
ex
=
+ 2x + C.
2
2
2
From the initial condition y(0) = −2, we see that (−2)
= 1 + 0 + C, and so C = 32 .
2 √ 2
√ 2
Consequently y = ± ex + 4x + 3. We reject y = ex2 + 4x + 3 because y(0) is
negative. So the solution is
p
y = − ex2 + 4x + 3.
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 6 of 12
Problems 3–6 are short-answer questions: put a box around your final answer, but you must include
the correct accompanying work to receive full credit.
Z
16x + 16
5a. [3 pts] Find
dx.
x(x − 4)(x2 + 4)
By the partial fraction of a rational function, we can write
16x + 16
A
B
Cx + D
= +
+ 2
,
2
x(x − 4)(x + 4)
x x−4
x +4
or equivalently
16x + 16 = A(x − 4)(x2 + 4) + Bx(x2 + 4) + (Cx + D)x(x − 4).
Plugging in x = 0, we see that A = −1; plugging in x = 4, we see that B = 1. Therefore
16x + 16 = −(x − 4)(x2 + 4) + x(x2 + 4) + (Cx + D)x(x − 4)
16x + 16 = (−x3 + 4x2 − 4x + 16) + (x3 + 4x) + (Cx + D)(x2 − 4x)
−4x2 + 16x = (Cx + D)(x2 − 4x),
from which we see that Cx + D = −4, and so C = 0 and D = −4. Therefore
Z
Z 16x + 16
−1
1
−4
dx =
+
+
dx
x(x − 4)(x2 + 4)
x
x − 4 x2 + 4
x
= − log |x| + log |x − 4| − 2 arctan + C.
2
5b. [3 pts] Find the sum of the series
∞ X
5
k=2
We can split this sum up as
∞ X
5
k=2
6
5
+ −
. Simplify your answer completely.
k 3k k + 1
6
5
+ k−
k 3
k+1
∞ ∞
X
X
5
5
6
+
−
=
,
k
3
k
k
+
1
k=2
k=2
as long as both of the new series converge. The first series is a geometric series with first
2/3
= 1. The second sum is a
term 362 = 32 and common ratio 13 , and so its sum is 1−1/3
telescoping series: its partial sums are
N X
5
5
5 5
5 5
5
5
5
5
−
=
−
+
−
+ ··· +
−
+
−
k k+1
2 3
3 4
N −1 N
N
N +1
k=2
=
5
5
−
,
2 N +1
and so its sum is
X
∞ N X
5
5
5
5
5
5
5
−
= lim
−
= lim
−
= − 0.
N →∞
N →∞
k k+1
k k+1
2 N +1
2
k=2
k=2
The sum of the original series is therefore 1 +
5
2
= 72 .
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 7 of 12
Problems 3–6 are short-answer questions: put a box around your final answer, but you must include
the correct accompanying work to receive full credit.
Z sin x
6a. [3 pts] Define F (x) =
log(3 + 2t) dt. Find a formula for the derivative F 0 (x).
cos x
u
Z
log(3 + 2t) dt, then we have
If we define g(u) =
0
Z
sin x
0
Z
log(3 + 2t) dt +
F (x) =
0
Z
log(3 + 2t) dt
cos x
Z cos x
sin x
log(3 + 2t) dt −
=
0
log(3 + 2t) dt
0
= g(sin x) − g(cos x).
By the Fundamental Theorem of Calculus, g 0 (u) = log(3 + 2u). Therefore we can find
F 0 (x) using the Chain Rule:
d
F 0 (x) =
g(sin x) − g(cos x)
dx
= g 0 (sin x)(cos x) − g 0 (cos x)(− sin x)
= log(3 + 2 sin x) cos x + log(3 + 2 cos x) sin x.
6b. [3 pts] Determine the interval of convergence of the power series
∞
X
(x − 3)k
k=1
k2k
.
First we compute
(x − 3)k+1 /(k + 1)2k+1 k 2k (x − 3)k+1 1 k =
k + 1 2k+1 (x − 3)k = 2 k + 1 |x − 3|,
(x − 3)k /k2k
and so
ak+1 (x − 3)k+1 /(k + 1)2k+1 1
= lim = · 1 · |x − 3|.
lim 2
k→∞ ak k→∞ (x − 3)k /k2k
By the Ratio Test, the series converges if 12 |x−3| < 1, that is, if |x−3| < 2 or equivalently
if 1 < x < 5; and it diverges if 21 |x − 3| > 1.
Next we examine the series’s behavior at the endpoints. If x = 5, then the series becomes
∞
∞
X
X
1
(−1)k
, which is a divergent p-series. If x = 1, then the series becomes
,
k
k
k=1
k=1
which converges by the Alternating Series Test. Therefore, the interval of convergence is
1 ≤ x < 5, or simply [1, 5).
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 8 of 12
Problems 7–11 are long-answer: give complete arguments and explanations for all your calculations—
answers without justifications will not be marked.
7. The two parts of this problem are unrelated.
(a) [3 pts] Calculate the volume of the solid obtained by rotating the region enclosed by
y = x2 and y = 2x about the y-axis.
An easy calculation finds that the graphs of y = x2 and y = 2x meet at the points (0, 0)
and (2, 4). Since we are rotating the region horizontally about the y-axis, we must solve
√
both equations for x in terms of y, yielding x = y and x = y2 . For the relevant values
√
of y, namely 0 ≤ y ≤ 4, we see that y ≥ y2 . Therefore the volume of the solid of
revolution is
2 Z 4
Z 4 y2
y
√ 2
y−
dy = π
dy
π ( y) −
2
4
0
0
2
4
y
y 3 16
8π
=π
−
=π 8−
.
=
2
12
3
3
0
(b) [4 pts] A circular wading pool has a diameter of 2 m and sides 20 cm tall. If it is half-full
of water, how much work (in joules) is required to pump all of the water out over the side?
(The density of water is 1000 kg/m3 . Use g = 9.8 m/s2 for the acceleration due to gravity.)
Let us choose coordinates so that y = 0 is the bottom of the pool. The thin horizontal
layer of water at height y(with thickness dy) must be lifted a total of (0.2 − y) meters
(since 20 cm = 0.2 m). The area of that cross-section is π · 12 m2 , and so its volume is
π dy m3 and its mass is 1000π dy kg; the force required to lift it is therefore 9.8 · 1000π dy
newtons. Hence the work required to lift that layer of water to the top of the pool is the
force times the distance, or 9800π dy · (0.2 − y).
To compute the total work required, we integrate this expression. The lowest horizontal
layer of water is at height y = 0; as the pool is only half-full, the highest horizontal layer
of water is at height y = 0.1. Therefore the appropriate integral is
0.1
Z 0.1
y 2 9800π(0.2 − y) dy = 9800π 0.2y −
2 0
0
= 9800π(0.02 − 0.005) = 147π joules.
Math 101—Practice Final Examination
Martin, Greg (#12345678)
d
dx
page 9 of 12
log(1 + sin x) = sec x − tan x
8. For both parts of this problem, you may use the formula
(you don’t have to show this).
(a) [5 pts] Let R be the region between the curves y = sec x and y = tan x, for x between 0
and π6 . Find the y-coordinate of the centroid of R.
sin x
Since 1 ≥ sin(x) and cos x > 0 for 0 ≤ x ≤ π6 , we have sec x = cos1 x ≥ cos
= tan x.
x
Thus the area of the region R is given by
Z π/6
π/6
(sec x − tan x) dx = log(1 + sin x)0
A=
0
π
3
= log 1 + sin
− log(1 + sin 0) = log ,
6
2
where we have used theFundamental Theorem of Calculus together with the given derivd
ative dx
log(1 + sin x) = sec x − tan x.
The y-coordinate of the centroid of R is then given by
Z
Z π/6
1 π/6 1
1
π/6
π
2
2
(sec x − tan x) dx =
1 dx =
=
,
A 0
2
2A 0
2A
12 log(3/2)
where we have used sec2 x − tan2 x = 1.
(b) [2 pts] Let S be the region between the curves y = sec x and y = | tan x|, for x between
− π6 and π6 . Using your answer to part (a), find the centroid of S.
Since both sec x and | tan x| are even functions of x, the region S is symmetric about the
y-axis. Therefore the centroid of S must lie on the y-axis, that is, the x-coordinate of the
centroid is zero. By the same symmetry, the y-coordinate of the centroid is the same as
that of the portion of S to the right of y-axis, whichis the region Rdiscussed in part (a).
π
Therefore, the coordinates of the centroid of S are 0,
.
12 log(3/2)
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 10 of 12
9. The two parts of this problem are unrelated to each other.
Z π
(a) [4 pts] Given that
10e−3x sin x dx = 1 + e−3π (you don’t have to show this), evaluate
0
Z π
−3x
10e
cos x dx.
0
Rπ
We canR use integration by parts to relate the integral 0 10e−3x cos x dx to the given inπ
tegral 0 10e−3x sin x dx = 1 + e−3π . With one choice of parts, say u = e−3x and
dv = cos x dx, we compute that
Z π
Z π
π
−3x
−3x
10e
cos x dx = 10e
sin x0 −
−3 · 10e−3x sin x dx
0
0
Z π
=0+3
10e−3x sin x dx
0
−3π
= 3(1 + e
) = 3 + 3e−3π
by the given integral.
The “other” choice of parts, namely u = cos x and dv = e−3x dx, also succeeds and gives
the same conclusion:
π Z π
Z π
10 −3x
10 −3x
−3x
e
cos x −
e (− sin x) dx
10e
cos x dx =
−3
0 −3
0
0
Z
1 π
10
−3π
=
10e−3x sin x dx
−e
−1 −
−3
3 0
10
1
= (1 + e−3π ) − (1 + e−3π ) = 3 + 3e−3π ,
3
3
as before.
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 10 of 12
(b) [3 pts] Does the improper integral
Z
π/2
F (x) =
π/3
sin t
√
dt
3
cos t
converge or diverge?
sin t
The integral is improper because the integrand √
has a vertical asymptote at t = π2 .
3
cos t
So we begin by writing
Z b
Z π/2
sin t
sin t
√
√
dt = lim −
dt .
3
3
b→π/2
cos t
cos t
π/3
π/3
Using the substitution u = cos t, so that du = − sin t dt, we find that
Z b
Z cos b
Z 1/2
sin t
1
1
√
dt = −
du =
du,
3
1/3
1/3
u
cos t
π/3
1/2
cos b u
since u = cos( π3 ) = 1/2 when t = π3 and u = cos b when t = b. By FTC,
1/2
Z 1/2
u2/3 3 1
1
2/3
du =
=
− (cos b)
,
1/3
2/3 cos b 2 22/3
cos b u
and therefore
Z b
3
sin t
1
3 1
2/3
√
dt =
lim
lim
− (cos b)
−0 ,
=
3
b→π/2−
2 b→π/2− 22/3
2 22/3
cos t
π/3
since (cos b)2/3 is continuous and equals 0 at b =
hence the improper integral does converge.
π
.
2
In particular, this limit exists, and
Math 101—Practice Final Examination
Martin, Greg (#12345678)
2
10. The Maclaurin series of the function √
4
page 11 of 12
6
3x
5x
1
x
+
+
+ · · · (you don’t have
is 1 +
2
2
8
16
1−x
to show this).
(a) [3 pts] Give the first four non-zero terms of the Maclaurin series for arcsin x.
First, note that
x
1
√
dt = arcsin t = arcsin x.
1 − t2
0
0
Therefore, integrating the given power series term-by-term gives
Z x
Z x
1
t2 3t4 5t6
√
1+ +
arcsin x =
dt =
+
+ · · · dt
2
8
16
1 − t2
0
0
x
t3 3t5
5t7
= t+ +
+
+ · · · 6
40
112
0
5
3
3x
5x7
x
+
+
+ ··· .
=x+
6
40
112
Z
x
(b) [4 pts] If f (x) = arcsin x, evaluate f (5) (0).
(5)
The coefficient of x5 in the Maclaurin series for arcsin x, by definition, is equal to f 5!(0) .
On the other hand, we worked out in part (a) that the coefficient of x5 in this Maclaurin
(5)
3
3
3
series is equal to 40
. Therefore f 5!(0) = 40
, or f (5) (0) = 40
(5!) = 9.
Alternately, differentiating once gives
1
x2 3x4 5x6
=1+
+
+
+ ··· ,
2
8
16
1 − x2
and differentiating this Maclaurin series four more times term-by-term gives
5
225 2
3
(6 · 5 · 4 · 3) x2 + · · · = 9 +
x + ··· ;
f (5) (x) = 0 + 0 + (4 · 3 · 2 · 1) x0 +
8
16
2
plugging in x = 0 also gives f (5) (0) = 9.
f 0 (x) = √
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 12 of 12
11.
1 1
1
1
1
(a) [3 pts] Show that the alternating series 1 − + −
+
−
+ · · · converges to a
4 9 16 25 36
value that is less than 0.9.
This series is the same as
∞
X
(−1)n−1
n=1
n2
. The terms of this series decrease in magnitude to
zero, so the Alternating Series Test applies, and S =
∞
X
(−1)n−1
n=1
the successive partial sums sN =
N
X
(−1)n−1
n=1
n2
n2
converges. Moreover,
alternate between overestimates of S and
(−1)N
;
(N + 1)2
when this term is negative (that is, when N is odd) we have S ≤ sN , while when this
term is positive (that is, when N is even) we have sN ≥ S. In particular, with N = 3 we
have S ≤ s3 . On the other hand, the partial sum
1 1
s3 = 1 − + = 1 − 0.25 + 0.1̄ = 0.861̄ < 0.9,
4 9
and so it follows that also S < 0.9, as was to be shown. (We could also have estimated
1 1
1 1
s3 = 1 − + < 1 − + = 0.875 < 0.9 to arrive at the same conclusion.)
4 9
4 8
underestimates of S. Indeed, the first term omitted from the N th partial sum is
Math 101—Practice Final Examination
Martin, Greg (#12345678)
page 12 of 12
1·3 1·3·5 1·3·5·7
(b) [4 pts] Show that the series 1 −
+
−
+ · · · converges absolutely.
3!
5!
7!
In other words, we need to show that the series
1 · 3 · · · (2n − 1)
1·3 1·3·5 1·3·5·7
+
+
+ ··· +
+ ···
1+
3!
5!
7!
(2n − 1)!
converges. We can apply the Ratio Test to this end: the (n + 1)st term of the series is
obtained from the nth by multiplying by 2n + 1 and dividing by 2n · (2n + 1), because
the numerator of the (n + 1)st term is the product of all the odd numbers up to 2n + 1 and
the denominator of the (n + 1)st term is the product of all the numbers, even and odd, up
to 2n + 1. In other words, if
1 · 3 · · · (2n − 1)
an =
,
(2n − 1)!
then
an+1 = lim 2n + 1 = lim 1 = 0.
L = lim n→∞
an n→∞ 2n(2n + 1) n→∞ 2n
1·3·5
1·3
+
+ ...
Since L < 1, we conclude from the ratio test that the series 1 +
3!
5!
converges.
An alternate approach to this problem is to begin with algebraically simplifying the expression an = 1·3···(2n−1)
for the nth term of the series. We can start to see how this works
(2n−1)!
by examining the first few terms of the series. Expanding the expressions 3! = 1 · 2 · 3,
5! = 1 · 2 · 3 · 4 · 5, and so on, we notice first that the series simplifies to
1
1
1
1+ +
+
+ ··· .
2 2·4 2·4·6
Then, factoring out powers of 2 in the denominators of these fractions, the series becomes
1
1 1
1
1
1
1
1
1+ + 2
+ 3
+ ··· = 1 + + 2
+ 3
+ ···
2 2 1·2 2 1·2·3
2 2 · 2! 2 · 3!
We can continue this pattern: writing (2n + 1)! = 1 · 2 · 3 · · · 2n · (2n + 1), we find that
1 · 3 · 5 · · · (2n − 1) · (2n + 1)
1
1
an+1 =
=
= n
.
1 · 2 · 3 · · · 2n · (2n + 1)
2 · 4 · 6 · · · (2n)
2 · n!
∞
X
1
We know quite a few ways to show that the series
converges; aside from
n
2 · n!
n=0
the Ratio Test, which works as before, a Comparison Test with either the convergent
∞
∞
X
X
1
1
geometric series
=
2
or
with
the
convergent
series
= e will do. Indeed, we
n
2
n!
n=0
n=0
∞
∞
X
X
1
xn
x
as
the
value
of
the
power
series
e
=
at x = 21 , so
can even recognize
n · n!
2
n!
n=0
n=0
that its exact value is e1/2 .