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Unformatted text preview: n! (4x − 1)n . 7. (−1, 1] 1. (−4, 4) 8. [−4, , 4] 2. converges only at x = 4 Explanation:
When
an 3. (−∞, ∞) 4 xn
=√,
n 11
4. − ,
44 then
√
n
xn+1
·n
=√
n+1 x
√
 x n
n
=  x
=√
n+1
n+1
an+1
an Thus
lim n→∞ an+1
an 5. converges only at x =
. n=1 6. [0, 8)
Explanation:
The given series has the form =  x . By the Ratio Test, therefore, the given series
converges when x < 1, and diverges when
x > 1.
We have still to check for convergence at
x = ±1. But when x = 1, the series reduces
to
∞
4
√
n 1
correct
4 ∞ n=1 an (x − a)n with
a n = n! 4n , a= 1
.
4 But if x = a,
lim n!4n (x − a)n = ∞ , n→∞ stiurca (mas7745) – 11.8: Power Series – meth – (91825) By the Ratio Test, therefore, the given series and so by the Divergence Test, the series (i) converges when x < 1, ∞
n n=1 n!4 (x − a) n (ii) diverges when x > 1. diverges whenever x = a. Consequently, the
given series
converges only at x =
010 1
.
4 which diverges by the Inegral Test. On the
other hand, at x = 1, the series becomes Find the interval of convergence of the series
∞
xn
.
(−1)n
2n + 5
n=1 1. interval of cgce = [−5, 2]
2. interval of cgce = [−1, 1) 4. converges only at x = 0
5. interval of cgce = (−1, 1] correct
6. interval of cgce = (−2, 5]
7. interval of cgce = [−1, 1]
8. interval of cgce = (−∞, ∞) Explanation:
When xn
,
2n + 5 then But
lim n→∞ 2n + 5
= 1,
2n + 7 in which case
lim n→∞ an+1
an =  x . ∞ n=1 (−1)n
2n + 5 which converges by the Alternating Series
Test.
Consequently, the
interval of convergence = (−1, 1] . 3. interval of cgce = (−1, 1) xn+1 2n + 5
2n + 5
an+1
=−
=  x
n
an
2n + 7 x
2n + 7 We have still to check what happens at the
endpoints x = ±1. At x = −1 the series
becomes
∞
1
2n + 5
n=1 10.0 points an = (−1)n 6 ....
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This document was uploaded on 09/17/2013.
 Summer '13

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