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APPM 1360 Exam #3 Solutions Fall 2009
- (16 pts) Do the following sequences converge or diverge? If convergent, give the limit. Explicitly show your reasoning.
(a) an = (−1)n^
n n + 1 (b)
(−1)n^
n n^2 + 1
n≥ 22 Solution: (a)Sequence diverges since
n n + 1 →^ 1 as^ n^ → ∞^ so^ an^ = (−1)
n n n + 1 oscillates between -1 and 1. (b)Sequence converges to 0 by Squeeze Theorem since,
−n n^2 + 1
nn n^2 + 1
≤ n n^2 + 1
for n ≥ 1
and (^) nlim→∞^ n n^2 + 1
= lim n→∞^ −n n^2 + 1
L′H
- (20 pts) For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges. Justify your answers.
(a)
∑^ ∞
n=
cos(nπ) n^3 /^2 − 1
(b)
∑^ ∞
n=
(−1)n ln(n) + n
(c)
∑^ ∞
n=
ne−n
Solution: (a) Series is absolutely convergent since
∑^ ∞
n=
∣∣^ cos(nπ) n^3 /^2 − 1
∑^ ∞
n=
∣∣^ (−1)
n n^3 /^2 − 1
∑^ ∞
n=
n^3 /^2 − 1
and
∑^ ∞
n=
n^3 /^2 − 1
converges by Limit Comparison Test with
∑^ ∞
n=
n^3 /^2
, a convergent p-series.
(b) Series is conditionally convergent since for n ≥ 1 we have ln(n) < n implies ln(n) + n < 2 n, and
∑^ ∞
n=
∣∣^ (−1)
n ln(n) + n
∑^ ∞
n=
ln(n) + n >
∑^ ∞
n=
2 n
so
∑^ ∞
n=
ln(n) + n diverges by Direct Comparison with
∑^ ∞
n=
2 n which is divergent (since it is a nonzero
multiple of a divergent harmonic series.) The alternating series
∑^ ∞
n=
(−1)n ln(n) + n converges by the Alter- nating Series Test since un = 1 ln(n) + n
and note that un+1 = 1 ln(n + 1) + (n + 1)
ln(n) + n
= un and (^) nlim→∞ un = 0. (c) Absolutely convergent by Ratio (or Root) test since,
nlim→∞
an+ an^ = lim^ n→∞
n + 1 en+1^ ·^
en n = lim^ n→∞
n + 1 n ·^
e =
e <^1
- (20 pts) Show all work.
(a) Use series to evaluate: lim x→ 0 sin(x)^ −^ tan
− (^1) (x) x^3. (b) Find the Taylor Polynomial of order 2 generated by f (x) = (1 − x)^1 /^2 at x = 0.
Solution: (a) Converges to 1/6 since,
xlim→ 0 sin(x)−tan
− (^1) (x) x^3 = lim x→ 0
“ x− x3!^3 + x5!^5 − x7!^7 +···
” −
“ x− x 33 + x 55 − x 77 +···
” x^3 , and moreover
xlim→ 0
“ x− x3!^3 + x5!^5 − x7!^7 +···
” −
“ x− x 33 + x 55 − x 77 +···
” x^3 = lim x→ 0
“ (^) x 3 3 −^ x
3 3!
”
“ (^) x 5 5! −^ x
5 5
”
“ (^) x 7 7 −^ x
7 7!
” +··· x^3 , and finally
xlim→ 0
“ (^) x 3 3 −^ x
3 3!
”
“ (^) x 5 5! −^ x
5 5
”
“ (^) x 7 7 −^ x
7 7!
” +··· x^3 = lim x→ 0
x^3
“ ( 13 − (^) 3!^1 )+
“ (^) x 2 5! −^ x
2 5
”
“ (^) x 4 7 −^ x
4 7!
” +···
” x^3 =^
1 3 −^
1 3! =^
1 6
(b) Note,
(1 − x)^1 /^2 = 1 +
∑^ ∞
k=
k
(−x)k^ = 1 −
x +
x^2 + · · · = 1 −
x 2 +
2! x
so P 2 (x) = 1 −
x 2 −^
x^2
(24 pts) Justify your answers.
(a) Find a Maclaurin Series for f (x) = xe−x/^2. (You may use your knowledge of the Maclaurin Series of ex^ to answer this question.)
(b) Use the first 3 non-zero terms of the series found in part (a) to approximate
0
xe−x/^2 dx.
(c) Estimate the error of the approximation found in part (b). (d) Is the approximation found in part (b) an underestimate or an overestimate?
Solution:
(a) xe−x/^2 = x ·
∑^ ∞
n=
(−x/2)n n! =^ x^ ·
∑^ ∞
n=
(−1)nxn 2 nn! =
∑^ ∞
n=
(−1)nxn+ 2 nn! =^ x^ −^
x^2 2 +^
x^3 22 2! −^
x^4 23 3! +^ · · ·
(b)
0
xe−x/^2 dx ≈
0
x − x
2 2
3 8
dx =
x^2 2
− x
3 6
4 32
1
0
=^1
+^1
(c)
0
xe−x/^2 dx =
2 −^
32 −^
5 · 23 · 3! +^ · · ·^ , so^ |Error| ≤
(One could also use the Taylor Remainder Term R 3 (x) to estimate the error.)
(d) Note, −
5 · 23 · 3! <^ 0 implies the approximation used in part (b) is an overestimate.
- (20 pts) Show all work.
(a) Find the interval of convergence of
∑^ ∞
n=
(n + 1)xn.
(b) For what values of x is the series conditionally convergent? absolutely convergent? divergent?
