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Answer Key for Exam 1
1(a)
∑^ ∞
n=
en^ is a geometric series with ratio r = (^1) e , which is between −1 and 1, so it will converge. Since the
first term is also a = (^1) e , it converges to
a 1 − r
1 e 1 − (^1) e
e − 1
A lot of people failed to recognize this. If we don’t care so much what the series converges to, we can look at ∫ (^) ∞
1
dx ex^
1
e−x^ dx = −e−x
1 =^ −^
ex
∞
1
e
e
If f (x) is a positive decreasing function for positive x then we have
∑^ ∞ n=
f (n) >
1
f (x) dx >
∑^ ∞
n=
f (n) >
2
f (x) dx >
∑^ ∞
n=
f (n) >
3
f (x) dx >...
e−x^ is such a function, so we can say
∑^ ∞ n=
en^
1
dx ex^
=^1
e
∑^ ∞
n=
en^
But
∑^ ∞
n=
en^
e
∑^ ∞
n=
en^ , so this tells us that
∑^ ∞
n=
en^ converges to a number between
e and
e
1(b) Since 1 ex^ + 3 is a positive decreasing function for all x, we have
1
dx ex^ + 3
∑^ ∞
n=
en^ + 3
∑^ ∞
n=
en^
e − 1
so
1
dx ex^ + 3 converges to a positive number which is less than
e − 1
. Or we can say, using a calculation
from above,
0 <
1
dx ex^ + 3
1
dx ex^
e
so
1
dx ex^ + 3 converges to a positive number which is less than
e
. One way to get the exact value is to
substitute u = ex^ + 3. Then du = ex^ dx = (u − 3) dx, or dx = (^) udu− 3 , and we have ∫ dx ex^ + 3
du u(u − 3)
and this can be looked up in the table or done as in problem 3. Another approach is ∫ (^) ∞
1
dx ex^ + 3
1
ex^ + 3 − ex ex^ + 3 dx
=
1
ex ex^ + 3
dx
=
(x − ln (ex^ + 3))
∞
This has the form ∞ − ∞, but we can fix this by using x = ln(ex):
∫ (^) ∞
1
dx ex^ + 3
=^1
(ln (ex) − ln (ex^ + 3))
∞
1
which is where we would be if we had substituted u = ex^ + 3. Now ∫ (^) ∞
1
dx ex^ + 3
=^1
ln
ex ex^ + 3
∞
1
ln
1 + 3e−x
∞
1 = −
ln
1 + 3e−x
∞
1
A shorter way to reach this point is to write ∫ (^) ∞
1
dx ex^ + 3
1
dx ex^ + 3
e−x e−x^
1
e−x^ dx 1 + 3e−x
and then substitute v = 1 + 3e−x. Since − dv 3 = e−x^ dx, this leads directly to ∫ (^) ∞
1
dx ex^ + 3
ln
1 + 3e−x
∞
1 = −
ln
1 + (^3) e
ln 1
=^1
ln
1 +^3
e
2(i) If we let w = x^2 then dw = 2x dx and 12 dw = x dx and we have
∫ x cos
x^2
dx =^1 2
cos w dw =^1 2 sin w + C = sin
x^2
+ C.
2(ii) If u =^1 x and dv = x cos
x^2
dx, then du = − dx x^2 and, by (i), v =^1 2 sin
x^2
, and we have ∫ cos
x^2
dx =
sin
x^2
2 x −
2 sin^
x^2
dx x^2
sin
x^2
2 x
∫ (^) sin (x 2 ) x^2 dx.
2(iii) Using the answer to (ii), we have
∫ (^) ∞
1
cos
x^2
dx =
sin
x^2
2 x
∞
1
1
sin
x^2
x^2 dx.
At ∞ the first term on the right has an infinite denominator and a numerator between −1 and 1, so it must be zero there, and it equals 12 sin 1, which is finite, at the lower endpoint x = 1. The remaining integral on the right must satisfy
−
dx x^2
sin
x^2
x^2 dx <
dx x^2
Since L’Hˆopital’s rule tells us that
xlim→∞
x + 1 x + 4
LH = lim x→∞
we finally have (^) ∫ (^) ∞
1
dx (x + 1)(x + 4)
ln 1 − ln
ln
Alternatively, we could do the integral by substituting u = x + 1 x + 4 , in which case
du = (x + 4) · 1 − (x + 1) · 1 (x + 4)^2 dx^ =^
3 dx (x + 4)^2.
When x = 1 we have u = 25 , and when x = ∞ we have u = 1 by the same L’Hˆopital’s rule calculation as before. So ∫ (^) ∞
1
dx (x + 1)(x + 4)
1
dx x+ x+4 (x^ + 4)^2 =^1 3
1
3 dx (x + 4)^2
x+ x+ =^1 3
(^25)
du u
=^1
ln |u|
1 (^25)
and as before this equals 1 3
ln 1 − ln
ln
4(a) We know that
∑^ ∞
n=
np^ converges if p > 1 and diverges if p ≤ 1, so p = 1 is the “last” p-series that diverges,
or, to put it another way, it is the p-series that diverges while having the largest possible denominator.
4(b) If we let u = ln x, then du = dx x , so ∫ dx x ln x
du u = ln |u| + C = ln | ln x| + C.
4(c) From 4(b) we have
∫ (^) ∞
3
dx x ln x = ln (ln x)|∞ 3 = ln (ln ∞) − ln (ln 3) = ∞.
Therefore
∑^ ∞
n=
n ln n must also be infinite. On the other hand, since ln x > 1 for x > e, and e is less than 3,
we have (^) ∞ ∑ n=
n ln n
∑^ ∞
n=
n
∑^ ∞
n=
n
so
∑^ ∞
n=
n ln n is an infinite series that diverges while being less than
∑^ ∞
n=
n
. Indeed there is no “smallest
infinite series that diverges”—given any divergent infinite series, it is always possible to find a smaller series that also diverges.
