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3-20 Find the radius of convergence and interval of convergence of the series.
โ โ
n=
x n
โ n
We will apply the ratio test.
x
n+
โ n + 1
n
x n
x
n โ n + 1
โ |x| as n โ โ.
Hence the radius of convergence is 1. For x = 1, the series is a divergent p-series, and for x = โ1, the series
is an alternating series, and since โ^1 n
is decreasing and converges to zero, the series converges. The interval
of convergence is therefore [โ 1 , 1).
โ โ
n=
nโ 1 x n
n 3
x
n+
(n + 1) 3
n
3
x n
xn
3
(n + 1) 3
โ |x| as n โ โ
Hence the radius of convergence is 1. For x = ยฑ1, the series converges absolutely and therefore converges.
Therefore the interval of convergence is [โ 1 , 1].
โ โ
n=
n
n x
n .
(n + 1)
n+ x
n+
nnxn
x(n + 1)
n+
nn
converges if and only if x = 0. Therefore the radius of convergence is 0 and the interval of convergence is
[0, 0].
โ โ
n=
n n 4
n x
n
(n + 1)
n+ x
n+
n 4 n x n
(4(n + 1)x
n
โ | 4 x| as n โ โ.
Therefore the radius of convergence is
1 4
. At the end points, x = ยฑ
1 4 , the sequence (โ1)
n n 4
n x
n diverges, so
its sum cannot converge. Therefore the interval of convergence is (โ
1 4
1 4
โ โ
n=
n x n
n ln n
x n+
n+ ln(n + 1)
n ln n
x n
x ln n
4 ln(n + 1)
x
โฃ as n โ โ.
Therefore the radius of convergence is 4. For x = 4, the sequence
โ โ
n=
n 1
ln n
satisfies the criteria for the alternating series test and hence converges. For x = โ4. the sequence
โ โ
n=
n
n
4 n^ ln n
โ โ
n=
ln n
diverges because for n โฅ 2,
1 n
1 ln n , and the harmonic series diverges. The interval of convergence is
therefore (โ 4 , 4].
โ โ
n=
n x 2 n
(2n)!
x 2 n+
(2n + 2)!
(2n)!
x 2 n
x 2
(2n + 1)(2n + 2)
โ 0 as n โ โ.
Therefore the radius of convergence is infinity and the interval of convergence is R.
โ โ
n=
n(x โ 1)
n
n + 1(x โ 1)
n+
โ n(x โ 1) n
n + 1(x โ 1) โ n
โ |x โ 1 | as n โ โ.
The series converes if |x โ 1 | < 1, so the radius of convergence is 1. If x = 0 or if x = 2, the series
diverges because
n(x โ 1)
n does not converge to zero. Therefore the interval of convergence is (0, 2).
โ โ
n=
(3x โ 2) n
n 3 n
x
2 n+
(n + 1)!(n + 2)!2^2 n+
n!(n + 1)!
2 n+
x^2 n+
x
2
4(n + 1)(n + 2)
โ 0 as n โ โ.
Therefore the domain of J 1 is R.
- The function A defined by
A(x) = 1 +
x
3
x
6
x
9
is called the Airy function after the English mathematician and astronomer Sir George Airy.
(a) Find the domain of the Airy function.
If we write A(x) =
n= anx 3 n , then we find that
an =
2 ยท 3 ยท 5 ยท 6 ยท... ยท (3n โ 1) ยท 3 n
Since (^) โฃ โฃ โฃ โฃ
x
3 n+ an+
x 3 n an
x
3
(3n โ 1) ยท 3 n
โ 0 as n โ โ,
the series converges for all values of x in R.
