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Calculus III
Math 143 Spring 2011
Professor Ben Richert
Exam 1
Solutions
Problem 1. (25pts) Consider the sequence
n^4 /^5
n=
(a–5pts) Write the first 3 terms in the sequence of partial sums associated to
n^4 /^5
n=
. (For this problem you do not need to write any English).
Solution.
24 /^5
24 /^5
34 /^5
(b–10pts) Decide if
n^4 /^5
n=
converges or diverges.
Solution. This sequence converges, that is, the limit of the sequence exists. In particular, (^) nlim→∞
n^4 /^5 = 0 because the numerator is constant and the denominator grows without bound.
(c–10pts) Compute the limit of the sequence of partial sums associated to
n^4 /^5
n=
Solution. By definition, the limit of the sequence of partial sums is
∑^ ∞
n=
n^4 /^5
=^4
∑^ ∞
n=
n^4 /^5 , which diverges by the P -test. Problem 2. (10pts) Do exactly one of the following two problems. Indicate clearly which one you want graded.
(a–10pts) Converge or Diverge:
∑^ ∞
n=
(2n + 1)n n^2 n
Solution. The root test states that a series
∑^ ∞
n=
an absolutely converges if (^) nlim→∞^ n
|an| < 1. Here
nlim→∞^ n
(2n + 1)n n^2 n^ = (^) nlim→∞^ n
(2n + 1)n (n^2 )n^ = (^) nlim→∞^ n
2 n + 1 n^2
)n = (^) nlim→∞ 2 n + 1 n^2 = (^) nlim→∞
n
n^2
So we conclude that
∑^ ∞
n=
(2n + 1)n n^2 n^ absolutely converges.^
(b–10pts) Converge or Diverge:
∑^ ∞
n=
n − 1 n 3 n
Solution. The comparison test states that if 0 ≤ an ≤ bn for all n ≥ 1 and
∑^ ∞
n=
bn converges, then
∑^ ∞
n=
an converges,
while the geometric series test states that
∑^ ∞
n=
arn−^1 converges if |r| < 1. Now note that 0 ≤ n − 1 ≤ n for all
n ≥ 1, and hence 0 ≤ n − 1 n 3 n^
n n 3 n^
3 n^
)n
. Thus by the comparison test it is enough to show that ∑^ ∞ n=
)n converges. But
∑^ ∞
n=
)n
∑^ ∞
n=
)n− 1 is geometric with |r| = | 1 / 3 | < 1 and hence converges by the geometric series test, so we are done. Problem 3. (10pts) Do exactly one of the following two problems. Indicate clearly which one you want graded.
(a–10pts) Converge or Diverge:
∑^ ∞
n=
n ln n Solution. We use the integral test, which states that if f (x) is positive, continuous and decreasing, and f (n) = an for all n ≥ 2, then
∑^ ∞
n=
an diverges if and only if
2
f (x) dx diverges.
Note that
x ln x is certainly positive for x ≥ 2, (^) xlim→∞
x ln x = 0 is obvious since x and ln x both grow without bound, and 1 x ln x is decreasing since both x and ln x are increasing (hence their product is increasing and their
reciprocal decreasing). So by the integral test,
∑^ ∞
n=
n ln n diverges if and only if
2
x ln x dx^ diverges. Now ∫ (^) ∞
2
x ln x dx = lim a→∞
∫ (^) a
2
x ln x dx,
and making the u-substitution u = ln x (whence the lie is that dx = du/u), we have
alim→∞
∫ (^) a
2
x ln x dx = lim a→∞
∫ (^) x=a
x=
u du
= lim a→∞ ln |u|
x=a x= = lim a→∞ ln(ln(x))
a 2 = lim a→∞ ln(ln(a)) − ln(ln(2)) = ∞.
We conclude that
∑^ ∞
n=
n ln n diverges.
(b–10pts) Converge or Diverge:
∑^ ∞
n=
n
(1/5)n
Solution. The limit comparison test states that if 0 ≤ an, bn and (^) nlim→∞ an bn = L > 0, then
∑^ ∞
n=
an and
∑^ ∞
n=
bn both diverge or both converge. So let an =
n
)n and bn =
)n , both of which are obviously positive, and note that
∑^ ∞
n=
)n
∑^ ∞
n=
)n− 1 converges by the Geometric Series Test. Thus it is enough to show that
nlim→∞
1 + (^1) n
(1/5)n (1/5)n^
= L > 0.
Here
nlim→∞
1 + (^1) n
(1/5)n (1/5)n^ = (^) nlim→∞
n
n^ lim→∞
n
= 1^4 = 1.
Note that we can pass the limit inside the continuous function x^4 by the series limit laws. Since the limit exists and is not zero, we conclude that
∑^ ∞
n=
1 +^1
n
(1/5)n^ converges by the limit comparison test.
Now if f (x) = sin x, then the (N + 1)st derivative is ± sin(x) or ± cos(x), each of which has absolute value bounded by 1 on the whole real line. Thus |f (N^ +1)(x)| ≤ 1 = M on all R and
|RN (1/2)| ≤
1 | 12 |N^ +
(N + 1)!
2 N^ +1^ · (N + 1)!
Now taking N = 4, we conclude that sin(1/2) ≈ T 4 (1/2) with error at most
. Of course,
T 4 (x) = 0 + x + 0x^2 −
x^3 + 0x^4 ,
so the required estimate is
sin(1/2) ≈ T 4 (1/2) =
