SAMPLE MIDTERM 2. MATH 216. SOLUTIONS.
Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit
SAMPLE MIDTERM 2. MATH 216. SOLUTIONS.
Problem 1. Let {xn } be a sequence in R such that there is θ e [0, 1) with
lxn+1 - xn l < θn
for all n e N. Show that {xn } is a fundamental sequence and therefore converges.
Solution. We know that for every n e N,
n
sn = θ k =
k=0
We also proved in class that lθln → 0 (for lθl < 1).
lim sn =
1 - θn+1
1 - θ .
Since lθn l = lθln , we have θn → 0, therefore
1
1 - θ .
Now fix an arbitrary ε > 0. Since {sn } is convergent, it is a fundamental sequence (by a theorem in the class). By the definition there exists a natural number Nε e N such that for all n, m 2 Nε we have lsn - sm l < ε. Let n, m 2 Nε and suppose without loss of generality that n > m.
Then
_ n _ n n
lxn - xm l = _ (xk - xk -1 ) _ < lxk - xk -1 l < θ k = sn - sm < ε.
_k=m+1 _ k=m+1 k=m+1
This proves that {xn } is fundamental (hence, by a theorem in the class, it is convergent). 口
Problem 2. a. Let {xn } and {yn } be two convergent sequences in R. Assume that there exists N e N such that for every n 2 N one has xn < yn . Show that
lim xn < lim yn .
b. Show that there exists two convergent sequences {xn } and {yn } in R such that for every n 2 1 one has xn < yn but
lim xn 2 lim yn .
Solution. a. We argue by contradiction. Denote x = limn二o xn , y = limn二o yn , and assume that x > y. Choose ε = (x - y)/2 > 0.
As limn二o xn = x, there is N1 e N, such that for all n 2 N1
lxn - xl < ε 年÷ x - ε < xn < x + ε,
in particular, x < xn + ε .
As limn二o yn = y, there is N2 e N, such that for all n 2 N2
lyn - yl < ε 年÷ y - ε < yn < y + ε
in particular, yn - ε < y .
Set N0 := max{N1 , N2 , N}. Then N0 2 N1 and N0 2 N2 . Therefore, x < xN0 +ε and yN0 - ε < y . This implies
x - y < xN0 + ε - (yN0 - ε) = xN0 - yN0 + 2ε = xN0 - yN0 + x - y.
Hence, xN0 - yN0 > 0, that is xN0 > yN0 . As N0 2 N, this contradicts to the assumption of the claim. Hence our assumption x > y was wrong, therefore x < y . This proves the statement.
b. Consider sequences defined by xn = 0 and yn = 1/n for every n. Clearly, xn < yn for every n e N and
0 = lim xn 2 lim yn = 0.
口
Problem 3. Let {xn } be a sequence in R, let x be an accumulation point of {xn }, and let {εk } be a sequence of strictly positive reals such that limk二o εk = 0. Show that there is a subsequence {xnk} of {xn } such that lxnk - xl < εk for all k e N.
Solution. For k e N denote
Nk = {n e N l lxn - xl < εk }.
As x is an accumulation point of {xn }, the set Nk is infinite for each k e N. We construct our subsequence inductively. Let n1 e N1 . Suppose that n1 < n2 < . . . < nk have already been chosen such that lxnj - xl < εj for j = 1, . . . , k. As Nk+1 is infinite, there is nk+1 e Nk+1 such that nk+1 > nk . As nk+1 e Nk+1, we have lxnk+l - xl < εk+1. Thus, the resulting subsequence {xn } has the desired property. 口
2022-02-15