Solution outlines for Stovall midterm 3
R
1. y 0 − 2y = x is a linear equation with P = −2x and Q = x, so P = −x2 . Therefore
Z
Z
R
R
1
1 −x2
2
− P
P
x2
−x2
x2
y=e
Qe = e
+ C = − + Cex
xe
dx = e
− e
2
2
(D)
n2
(−1)n n2
=
0,
we
have
lim
= 0 as well
n→∞ 1 + n3
n→∞ 1 + n3
2. Since lim
(A)
3. Can use the integral test (with the substitution u = 1 + x2 ):
Z ∞
Z
√ ∞
x
1 ∞ −1/2
√
u
du = u = ∞.
, dx =
2 2
1 + x2
1
2
Since the integral diverges (and is infinite), the series diverges and the sum is infinite.
(B)
4. For sufficiently large n, we know that ln n < n0.1 Therefore
X ln n 2 X (n0.1 )2 X 1
<
=
n
n2
n1.8
which is a convergent p-series with p = 1.8 > 1. So the original series converges by the
direct comparison test.
X 3 + cos n X 4
5. I.
<
, which is a convergent geometric series with r = 13 < 1. So
3n
3n
the original series converges by direct comparison.
II. Use the ratio test (and divide the numerator and denominator by 6n = 2n · 3n ):
1
1
1
1
+2
+1
(1 + 2n+1 ) n (1 + 3n ) n
1 + 2n+1 1 + 3n
2
2n
3n
2
3
n
= <1
lim
·
= lim
= lim n+1
n
1
1
n→∞ 1 + 3
n→∞
n→∞
1
1
1+2
3
(1 + 3n+1 ) n (1 + 2n ) n
+3
+1
n
3
2
3
2
therefore the series converges.
So both series converge.
(D)
6. I. This series is not alternating (so can’t be conditionally convergent). Ratio test:
(n + 3)(n + 2)
1
10n
=
<1
·
n+1
n→∞
10
(n + 2)(n + 1)
10
so this series converges absolutely.
n
is decreasing and its limit as n → ∞ is zero, the series converges by
II. Since 2
n +4
the alternating series test. But the integral:
∞
Z ∞
x
1
2
=∞
dx
=
ln(x
+
4)
x2 + 4
2
1
1
X n
of absolute values diverges and so the original series converges
so the series
n2 + 4
conditionally.
(E)
lim
7. First use the ratio test:
n+1 n+1
3 x
(n + 1)2 lim
= 3|x|
·
n→∞ (n + 2)2
3n xn For the ratio test to guarantee convergence we need 3|x| < 1, so − 13 < x < 13 .
X
X 1
1
Next check the endpoints: At x = 31 the series becomes
<
(a
(n + 1)2
n2
convergent p-series), so the series converges at x = 31 . Likewise at x = − 13 , the same
test shows the series converges absolutely.
So the interval of convergence is [− 31 , 13 ] and the radius of convergence is r = 13
(B)
8. Since cos Z =
∞
X
−1)n Z 2n
n=0
(2n)!
with infinite radius of convergence, we can replace Z by
x3 to get
3
cos x =
∞
X
−1)n (x3 )2n
n=0
(2n)!
again with infinite radius of convergence.
(C)
=
∞
X
−1)n x6n
n=0
(2n)!