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Problem set 3 for mathematics 120, focusing on the convergence and divergence of sequences and series. Students are required to determine if given sequences converge or diverge, find their limits, and decide if specific series converge or diverge. The document also includes instructions for using the maclaurin series to prove that e is irrational.
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Mathematics 120 Problem Set 3 Part A ( Problems 1 - 7)
2 −^ n^1
(b)
3 −n
(c)
{ (^7) n (^3) + 16n − 11 4 n^3 + 26
(d)
an
, given that a 1 = 5 and an+1 = − 0. 3 an for n ≥ 1
(e)
ln
( (^) n n + 1
(f)
n + 1 −
n
(g)
{ (^100) n n!
(a)
n=
n^1
(b)
n=
5 −n
(c)
n=
n^4 + 6n^2 + 10 3 n^4 + 27
(d)
n=
an , given that a 1 = 5 and an+1 = − 0. 3 an for n ≥ 1
(e)
n=
ln
( (^) n n + 1
(a)
n=
n + 1
n
(b)
n=
100 n+ n!
(c) 1 2002
(d)
n=
n + 17
)n
(e)
n=
ln 2n^2
Math 120 Prob Set 3A continued
A = (q + 1)! [e − 1 − 1 −
(q + 1)!
(a) Explain how we know that
e − 1 − 1 −
(q + 1)!
(q + 2)!
q + 3)!
(b) Using part (a), explain how we know that A is positive. (c) Using the hypothesis relating p, q and e, explain how we know that A is an integer. (d) Explain the steps in the following calculation.
A = (q + 1)![e − 1 − 1 − 1 2!
(q + 1)!
q + 2
(q + 2)(q + 3)
q + 2
(q + 2)^2
q + 2
q + 2
(q + 2)^2
q + 2
1 − (^) q+2^1 = 1 q + 1
(e) Parts (b) and(c) tell us that A is a positive integer. Part (d) tells us that A <
q + 1. Explain why these results contradict each other. (This contradiction shows that our assumption must be false. In other words, e must be irrational.) (f) Now that you’ve completed the proof, summarize it as concisely as you can. Your summary can be a prose summary or a summary outline.
n=
an is a convergent series with nonnegative terms.
(a) Explain how we know that there is a positive integer N such that an < 1 provided n > N.
(b) With N as in part (a), use the Direct Comparison Theorem to show that
n=N
a^2 n converges.
(c) Use part (b) to show that
n=
a^2 n also converges.
(d) Now give another proof that
n=
a^2 n converges, this time basing your argument on the Limit Com- parison Test.