Math 113 (Calculus Ii) Test 4 Form A With Answer Key - Brigham Young University Page 4
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1
2
3
4
5and determine its radius of convergence.
Solution: Since
1
1
dx = tan
(x) + C
2
1 + x
and
1
n
2n
dx =
( 1)
x
,
1 + x
2
n=0
then
n
( 1)
1
2n+1
tan
(x) =
x
.
2n + 1
n=0
(The constant is 0 because inverse tangent is 0 at 0). Thus,
n
( 1)
1
2n+2
x tan
(x) =
x
.
2n + 1
n=0
2
12. (8 points) Find the Maclaurin series for x
sin 2x and give its radius of convergence.
Solution: The Maclaurin series for sin(2x) is given by
n
2n+1
( 1)
2
2n+1
sin(2x) =
x
,
(2n + 1)!
n=0
so
n
2n+1
( 1)
2
2
2n+3
x
sin 2x =
x
.
(2n + 1)!
n=0
n
13. (8 points) Test the series
for convergence or divergence.
(ln n)
n=2
Solution: You can use the integral test or the comparison test or the divergence test. Notice
by L’Hopital’s rule that
x
1
lim
= lim
= lim
x =
.
1
ln x
x
x
x
x
So, by the divergence test, the series diverges.
n
( 1)
14. (10 points) Show that the series
is convergent. How many terms of the series are
n
n=1
needed to find the sum to an accuracy of |error| < 0.01? Explain your answer.
1
1
Solution: Notice that
n <
n + 1, so
>
. Thus, the sequence
n
n + 1
1
{
}
n
is decreasing and clearly goes to 0. Hence, the series converges by the alternating series test.
To estimate the error, we want the next term out to be smaller than 0.01, or
1
< 0.01
n + 1
or
n + 1 > 100, n + 1 > 10000, n > 9999.
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