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Unformatted text preview: when x > 1. x→∞ 005 n=1 √ 1
.
n+5 Applying the Integral Test with ∞ cn 7n
n=0 is convergent, which of the following statements must be true without further restrictions on cn ?
∞ 1
f ( x) = √
x+5 1. we see that f is continuous, positive, and decreasing on [1, ∞), but the improper integral 2. n=0 n=0 f (x) dx ∞ (‡) n=1 (−1)n
√
.
n+5 cn (−3)n is convergent correct ∞ 1 diverges, so the inﬁnite series (∗) diverges
also.
On the other hand, at x = −1, the series
becomes cn (−3)n is divergent ∞ ∞ I= 10.0 points If the series We have still to check what happens at the
endpoints x = ±1. At x = 1 the series
becomes
( ∗) 1
= 0,
x+5 however, the Alternating Series Test ensures
that (‡) converges.
Consequently, the so ∞ 1
x+5 the same continuous, positive and decreasing
function as before. Since an+1
an lim a n = f ( n) with xn
=√
,
n+5 √
xn+1
n+5
=√
n + 6 xn
√
n+5
=  x
=  x √
n+6 (−1)n an , 3.
n=0 cn (−7)n is convergent ∞ 4.
n=0 cn (−7)n is divergent Explanation: stiurca (mas7745) – 11.8: Power Series – meth – (91825)
The series with ∞ cn xn an = n=0 converges when x = 7, its radius of convergence, R, must have the property R ≥ 7 since
the series
(i) converges when x < R, and
(ii) diverges when x > R. Consequently, it must be true that the series
∞ n=0 cn (−3) n is convergent but we cannot determine whether either of
the statements
∞ (i)
n=0
∞ cn (−7)n is convergent,
n (ii)
n=0 cn (−7) is divergent For the series
∞ xn n=1 (n + 6)! , (i) determine its radius of convergence, R. 1
.
(n + 6)! Now for this series,
(i) R = 0 if it converges only at x = 0,
(ii) R = ∞ if it converges for all x,
while 0 < R < ∞,
(iii) if it converges when x < R, and
(iv) diverges when x > R.
But
lim n→∞ an+1
an = lim n→∞ 1
= 0.
n+7 By the Ratio Test, therefore, the given series
converges for all x. Consequently,
R=∞ . must be true without further conditions on
cn .
006 (part 1 of 2) 10.0 points 4 007 (part 2 of 2) 10.0 points
(ii) Determine the interval of convergence
of the series.
1. (−6, 6) 1. R = 0 2. [−1, 1] 2. R = 6 3. (−∞, ∞) correct
4. converges only at x = 0 3. R = 1
5. [−6, 6] 1
4. R =
6 6. (−1, 1] 5. R = ∞ correct Explanation:
The given series has the form Explanation:
Since R = ∞, the given series converges for
all x, and so has ∞ a n xn
n=1 interval convergence = (−∞, ∞) . stiurca (mas7745) – 11.8: Power Series – meth – (91825)
008 10.0 points Find the interval of convergence of the
power series
∞
4 xn
√.
n which diverges by the pseries test with p =
1
≤ 1. On the other hand, when x = −1, the
2
series reduces to
∞ n=1 1. [−1, 1) correct 5 n=1 4
(−1)n √
n which converges by the Alternating Series
Test. Thus the 2. (−1, 1) interval of convergence = [−1, 1) . 3. (−4, 4) 009
4. [−4, 4) 10.0 points Determine the interval of convergence of
the series of the series 5. (−4, 4] ∞ 6. [−1, 1] n=1...
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 Summer '13

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