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Calculus III
Math 143 Fall 2010
Professor Ben Richert
Exam 1
Solutions
Problem 1. (25pts) Consider the sequence
n n + 1
n=
(a–10pts) Does
n n + 1
n=
converge or diverge?
Solution. In this case the sequence converges because (^) nlim→∞ n n + 1 exists. Using the usual nerd trick and the limit laws, we find
nlim→∞
n n + 1 = (^) nlim→∞
1 n 1 n
n n + 1
= (^) nlim→∞
1 + (^) n^1
(b–10pts) Does the sequence of partial sums of
n n + 1
n=
converge or diverge?
Solution. Given a sequence {an}∞ n=0, we use
∑^ ∞
n=
an to denote the limit of the sequence of partial sums (when it
exits). So this question is asking if
∑^ ∞
n=
n n + 1 converges or diverges. Since^
n n + 1 6 →^ 0 (as we saw in part (a)), we
conclude that
∑^ ∞
n=
n n + 1 diverges by the Test for Divergence.^
(c–5pts) Use your answer in part (b) to show that the power series
∑^ ∞
n=
n n + 1 xn^ has radius of convergence ≤ 1.
Solution. If the radius of convergence of this power series was R > 1, then the series would converge for all |x| < R, and in particular, it would converge for x = 1. This cannot be the case, however, since we found in part (b) that ∑^ ∞
n=
n n + 1
∑^ ∞
n=
n n + 1 1 n^ diverges. It must be, therefore, that the radius of convergence is ≤ 1.
Problem 2. (10pts) Do exactly one of the following two problems. Indicate clearly which one you want graded.
(a–10pts) Decide whether
∑^ ∞
n=
(−1)n 6 n^ + 1 converges or diverges.
Solution. It is clear that
∑^ ∞
n=
(−1)n 6 n^ + 1 alternates. In addition (^) nlim→∞
6 n^ + 1 = 0 (because the denominator 6n^ + 1
increases without bound as n → ∞ and therefore
6 n^ + 1 becomes arbitrarily small), and
6 n^ + 1 is decreasing
(since 6n^ + 1 is an increasing function). Thus
∑^ ∞
n=
(−1)n 6 n^ + 1 satisfies the conditions for the Alternating Series Test, and hence converges.
(b–10pts) Decide whether
∑^ ∞
n=
n + 1 2 n + 2
)n converges or diverges.
Solution. We use the Root Test. Note that
nlim→∞^ n
n + 1 2 n + 2
)n∣∣ ∣∣ = lim n→∞
n + 1 2 n + 2 = (^) nlim→∞
1 n 1 n
n + 1 2 n + 2
= (^) nlim→∞ 1 + (^) n^1 2 + (^) n^2
Since the limit is < 1, we know that
∑^ ∞
n=
n + 1 2 n + 2
)n converges absolutely by the Root Test.
Problem 3. (10pts) Do exactly one of the following two problems. Indicate clearly which one you want graded.
(a–10pts) Decide whether
∑^ ∞
n=
n
ln n
converges or diverges.
Solution. Consider the function f (x) =
x
ln x
. We know that f (x) > 0 for x > 1, f (x) is continuous for x > 1 (being made of continuous function which are never zero for x > 1), and f (x) is decreasing for x > 1 (since both x and
ln x are increasing functions (the latter being the composition of two increasing functions)). Also, we see that f (n) =
n
ln n
for all n ∈ N. Thus, by the integral test,
∑^ ∞
n=
n
ln n
diverges if and only if
2
x
ln x
dx diverges. So here we go: (^) ∫ ∞ 2
x
ln x
dx = lim b→∞
∫ (^) b
2
x
ln x
dx = lim b→∞
∫ (^) ln b
ln 2
u^1 /^2 dx
= lim b→∞ 2 u^1 /^2
ln b ln 2 = lim b→∞
ln b − 2
ln 2) = ∞. We conclude that the integral (and hence also the series) diverges. Here we used the u-substitution u = ln x (whence du dx =
x , or lying,^ du^ =^
dx x ) and changed the limits of integration to reflect that^ u^ = ln 2 when^ x^ = 2 and u = ln b when x = b. We also used that fact that lim b→∞
ln b = ∞.
(b–10pts) Decide whether
∑^ ∞
n=
3 n^2 + 1 n^2 + 2
en^ converges or diverges.
Solution. We compare with
∑^ ∞
n=
en^ using the Limit Comparison Test. Note that
nlim→∞
3 n^2 + n^2 +
1 en 1 en
= (^) nlim→∞
3 n^2 + 1 n^2 + 2
= (^) nlim→∞
1 n^2 1 n^2
3 n^2 + 1 n^2 + 2
= (^) nlim→∞ 3 + (^) n^12 1 + (^) n^22
so by the Limit Comparison Test, both
∑^ ∞
n=
3 n^2 + 1 n^2 + 2
en^ and
∑^ ∞
n=
en^ converge or both diverge. But
∑^ ∞ n=
en^
∑^ ∞
n=
e
e
)n− 1
is a geometric series with r =
e < 1 and is hence convergent by the Geometric Series Test. We conclude that ∑^ ∞ n=
3 n^2 + 1 n^2 + 2
en^ converges as well.
Problem 4. (10pts) Compute T 3 (x), the Taylor polynomial of degree 3 for the function f (x) = sin(x) cos(x). Solution. The formula for the Taylor polynomial of degree 3 (centered at x = 0) is
T 3 (x) =
∑^3
n=
f (n)(0) n! xn^ = f (0)(0) + f (1)(0)x + f^
x^2 + f^
x^3.
So, plugging and chugging, we have:
on |x| < 1. In particular,
|R 4 (1/3)| ≤
3(1/3)^5
so we conclude that the estimate
e^1 /^3 ≈
∑^4
n=
(1/3)n n!
has error at most
as required.
