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The solutions to exam 4 of math 2202, focusing on topics such as the ratio test for series convergence, finding the radius of convergence of power series, obtaining the taylor series of a function, and approximating definite integrals using maclaurin series. Students can use this document as a valuable resource for understanding these concepts and preparing for exams.
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MATH 2202 - Exam 4 (Version 1) Solutions November 23, 2009 S. F. Ellermeyer Name
Instructions. Show all of your work! You will not get full credit if you donít include correct notation. In particular, you must write ì=îwhere needed. Also, when discussing sequences and series, take care not to write the word ìitî. Special Thanksgiving Bonus!: There are six questions on this exam but I will only use your best Öve to compute your grade.
n=
p n 1 + n^2
converges or diverges. Does the Ratio Test succeed of fail? If the test succeeds, then does it tell you that the series converges or diverges (and why)? If the test fails, then why does it fail? Solution: Since an =
p n 1 + n^2
then an+1 =
p n + 1 1 + (n + 1)^2 and an+ an
p n + 1 1 + (n + 1)^2
1 + n^2 p n
=
r n + 1 n
n^2 + 1 n^2 + 2n + 2
Since
n^ lim!
an+ an = lim n!
r n + 1 n
n^2 + 1 n^2 + 2n + 2
p 1 1 = 1,
the Ratio Test fails to give any information about the convergence or divergence of the series. (The series can be shown to converge by the Limit Comparison Test.)
n=
n^2 xn 2 4 6 (2n)
Solution: We have bn = n^2 xn 2 4 6 (2n)
and bn+1 = (n + 1)^2 xn+ 2 4 6 (2n) (2n + 2)
Thus
bn+ bn
(n + 1)^2 xn+ 2 4 6 (2n) (2n + 2)
2 4 6 (2n) n^2 xn
=
n + 1 n
2 n + 2
jxj
and we see that
n^ lim!
bn+ bn = lim n!
n + 1 n
2 n + 2
jxj = (1)^2 (0) jxj = 0.
By the Ratio Test, we conclude that the given series converges no matter what x is. In other words, the radius of convergence is 1.
1 1 x = 1 + x + x^2 + x^3 + x^4 + x^5 +
which holds for all x 2 ( 1 ; 1) and use substitutions and algebra to obtain the power series representation of x^2 1 + 9x^2
What is the interval of convergence of the power series that you obtain? (Be detailed in your solution.) Solution: Note that x^2 1 + 9x^2 = x^2
1 ( 9 x^2 )
Since 1 1 x
n=
xn^ for all x such that jxj < 1 ,
then 1 1 ( 9 x^2 )
n=