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Unformatted text preview: exp t log 1
and so all we have to show is that if we define x
t
for t x then the function g is increasing. Since g x 0 and since g is continuous on the interval
0 whenever t x. Now
x, Ý it will follow that g is increasing when we have shown that g t
whenever t x we have
g t t log 1 g t log 1 x
t and therefore what we want to show is that the inequality
log 1 u u
1u
holds whenever 0 u 1. We define
h u log 1 u x
t 1 x
t 0 u
1u
and observe that h 0 0 and that whenever 0 u 1 we have
u
1 1 u u
hu
1u
1 u2
u1
Therefore h u 0 whenever 0 u 1 and our proof is complete. 2 b. Use the dominated convergence theorem to show that
n
Ý
xn
1 n x 1 Þ e x x 1 dx
lim Þ
n Ý 1/n
0 0 . For the definition of the gamma function , see an earlier exercise. Some Exercises on Power Series
1. Find the radius of convergence and the interval of convergence of each of the following series:
a. n! 2 x n
2n !
An easy application of d’Alembert’s test shows that the series converges when x  4. To see
what happens when x 4 we define
n! 2 4 n
an
2n !
for each n. Since
n1 ! 2 4 n1
2 n 2 !
2n 2n 2 1
2n 1
n! 4
2n !
we see that a n does not approach 0 as n Ý and so the given series must diverge when
x 4. Thus the interval of convergence is 4, 4 and the radius of convergence is 4.
a n 1
an b. 2n ! x n
An easy application of d’Alembert’s test shows that the series converges when
n! 2
x  1 . To see what happens when x 1 we define
4
4
2n !
an n
4 n! 2
for each n. Since 372 2 n 2 !
4 n1 n1 ! 2 1 2n 1 2n 1 1
2 n1
2n 2
2n !
n n! 2
4
and so the sequence a n is decreasing. Since
n
lim n 1 aa 1 1 1
nÝ
n
2
Raabe’s test guarantees that a n is divergent but the criterion for terms to approach zero
guarantees that a n 0 as n Ý. It therefore follows from Dirichlet’s testDirichlet’s test that,
1 n a n is convergent. The interval of convergence of
although
a n is divergent, the series
n
2n ! x
11
is therefore
,
.
the series
44
n! 2
a n 1
an c. n! x n
nn
An easy application of d’Alembert’s test shows that the series converges when x  e. To see
what happens when x e we define
n! e n
an
nn
for each n. Since
n 1 ! e n 1
n 1 n 1
n
lim n 1 aa 1 n Ý n 1
lim
nÝ
n
n! e n
nn
e
1 1
n n Ýn 1
lim n 1
2 we know that the sequence a n does not approach zero and
a n diverges. When x e we
have
n! x n
n! e n
1n
n
n
nn
and, once again, because this nth term fails to approach zero the series diverges. The interval
n! x n
of convergence of the series
is therefore e, e .
nn
nnxn
d.
n!
We saw in an earlier group of exercises that the interval of convergence of this series is
11
e, e .
2. Given that c 1 and that
1
1x
1, expand the function f in a power series with center c and find the interval of convergence of
fx whenever x
this series. Hint: Use the identity
1
x 1 x 1 1 1
c 1 xc
1c . Thus
1
1
The series xc
1c n Ý 1
1 converges if and only if 373 c n0 x
1 c
c n x
1...
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 Fall '08
 STAFF
 Math, Calculus

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