Unformatted text preview: st with a pseries, p > 1.
iv) Diverges: Ratio test.
v) Diverges: Comparison or limit comparison with the harmonic series.
a) Answer: i) b) Answer: v) 3.6 Determine if the following series converge or diverge. Give your reasoning
using complete sentences.
a) ∞
n=1 b) (−1)n ∞
n=2 n
n+2 (−1)n+1 1
ln n a) An alternating series, but the terms do not approach 0 (they approach 1)
so divergent.
b) An alternating series with terms decreasing to 0 so convergent.
3.7 a) Find the interval of convergence for the power series
b) Find the radius of convergence for ∞ n! n
x
n=1 (2n)! ∞ 1
(x − 3)n
n
n=1 n2 MTH 142 Exam 3 Spr 2011 Practice Problem Solutions 4 a)
1
(x − 3)n+1
(n+1)2n+1
1
(x − 3)n
n2n = n2n (x − 3)n+1
1
n
1
=·
x − 3 → x − 3
n+1 (x − 3)n
(n + 1)2
2 n+1
2 as n → ∞. 1 x − 3 < 1 for x − 3 < 2, that is, 1 < x < 5 so series converges
2
if 1 < x < 5 and diverges if x > 5 or x < 1. At x = 5 the series becomes the
1
harmonic series n n which diverges, but at x = 1 it becomes the alternating
1
series n (−1)n n which converges. The interval of convergence is therefore
[1, 5).
b)
(n+1)! n+1
x
(2n+2)!
n!
xn
(2n)! = n+1
x → 0 &...
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 Fall '08
 POOLE
 Calculus, Cos, Mathematical Series, dx, dθ

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