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Problem 1. Find the radius R and interval I of convergence for the following series. Show
your work. Each series has the form
X
an (x − x0 )n .
n
(a) The series is a geometric progression series with the ratio
It means that R = 2, I = (3 − 2; 3 + 2) = (1; 5).
(b) The radius of convergence
q
√
n
3n n = 3.
R = lim
x−3 2
.
2
It converges iff
n→+∞
The endpoints: if x = −5, then the series
n
X 1
X
n−1 (−3)
√
√
=
−
(−1)
n n
3
n
n
n
diverges because 1/2 ≤ 1 (by the integral test); if x = 1, then the series
X
X (−1)n
3n
√
(−1)n−1 n √ = −
3 n
n
n
n
1
is alternate and converges by the alternate series test, √n+1
< √1n .
So, I = (−5; 1].
(c) The radius of convergence (use the Stirling formula)
s
r
n
3
10e n n5/2
n 10 n
√ = 0.
R = lim
= lim
n→+∞
n→+∞ n
n!
2π
The series converges (to 0) iff x = −2.
2. Find the function f (x) whose power series is given in each problem below.
(a) We have
+∞
d X n
d 1
x
f (x) = x
x =x
=
.
dx n=0
dx 1 − x
(1 − x)2
(b) We have
!
Z
+∞
+∞ Z
+∞
X
xn
1X x n
1 x X n
f (x) =
=
t dt =
t dt
n+1
x n=1 0
x 0
n=1
n=1
Z
Z
1
−x − ln(1 − x)
1 x t
1 x
=
dt =
...