McGill University
Department of Mathematics and Statistics
MATH 254 Analysis 1, Fall 2015
Assignment 5: Solutions
1. Let (yn ) be an unbounded sequence of positive numbers satisfying yn+1 > yn for all n ∈ N. Let
(xn ) be another sequence and suppose that the limit
lim
exists. Prove that
lim
xn+1 − xn
yn+1 − yn
xn
xn+1 − xn
= lim
.
yn
yn+1 − yn
Hint: You may use the Problem 3 on Assignment 4.
Using the above result, prove that for any p ∈ N the following holds:
(a)
lim
1p + 2p + · · · + np
1
=
np+1
p+1
(b)
lim
1p + 2 p + · · · + np
n
−
p
n
p+1
=
1
2
(c)
lim
1p + 3p · · · + (2n + 1)p
2p
=
np+1
p+1
Solution:
n+1 −xn
Let lim xyn+1
−yn = L. Then, for any > 0, ∃ N ∈ N such that for n ≥ N ,
xn+1 − xn
xn+1 − xn
yn+1 − yn − L < ⇐⇒ L − < yn+1 − yn < L + .
Using that (yn ) is a strictly increasing sequence,
(L − )(yn+1 − yn ) < xn+1 − xn < (L + )(yn+1 − yn ),
∀n ≥ N
Summing from some n0 ≥ N to some m ≥ n0 + 1,
(L − )
m−1
X
n=n0
(yn+1 − yn ) <
m−1
X
n=n0
xn+1 − xn < (L + )
m−1
X
(yn+1 − yn ).
n=n0
Since we have a telescoping sum, the previous equation is equivalent to:
(L − )(ym − yn0 ) < xm − xn0 < (L + )(ym − yn0 )
Since (yn ) is unbounded, we can choose m large enough so that ym is positive; then we may
divide by ym without changing the sides of the inequalities:
xm xn0
yn0
yn0
<
−
< (L + ) 1 −
(L − ) 1 −
ym
ym
ym
ym
Since (yn ) is unbounded, taking the limit as m → ∞ gives us:
L − ≤ lim
m→∞
xm
≤ L + .
ym
Since is an arbitrary positive number, we have (after replacing the dummy m by n):
lim
xn+1 − xn
xn
= L = lim
yn
yn+1 − yn
as we wanted to show.
(a) We want to compute
lim
(n + 1)p
(n + 1)p+1 − np+1
if it exists. Expanding,
(n + 1)p
(n + 1)p+1 − np+1
Pp
p
p
p−k
n−k
1 + k=1
n
+ k=1
k
k
=
=
Pp+1 p + 1
Pp+1 p + 1
p−k
p
n
n−k
(p + 1)n + k=2
(p + 1) + k=2
k
k
Pp
np
Taking the limit as n → ∞,
lim
(n + 1)p
1
=
p+1
p+1
(n + 1)
−n
p+1
lim
1p + 2p + · · · + np
1
=
p+1
n
p+1
By the theorem above,
(b) Let
1p + 2p + · · · + np
n
−
=
np
p+1
Then
Pn
k=1 k
np
p
P
(p + 1) nk=1 k p − np+1
n
xn
−
=
=
.
p+1
np (p + 1)
yn
xn+1 − xn
(p + 1)(n + 1)p − (n + 1)p+1 + np+1
=
.
yn+1 − yn
(p + 1) · [(n + 1)p − np ]
Using binomial theorem, the above expands into
Pp
Pp
p
p+1
p+1
p
p−1
p−k
p
p−1
p+1−k
(p + 1) · n + pn
+ k=2
n
− (p + 1)n +
n
+ k=3
n
k
2
k
Pp
p
p−1
p−k
(p + 1) · pn
+ k=2
n
k
2
Cancelling the np terms and regrouping the np−1 terms, we get
h
(p + 1)p −
(p+1)!
(p−1)!2!
i
P
P
p
p+1
np−1 + (p + 1) · pk=2
np−k − pk=3
np+1−k
k
k
.
Pp
p
p−1
p−k
(p + 1) · pn
+ k=2
n
k
Dividing top and bottom by np−1 , we get
(p + 1)p −
(p+1)!
(p−1)!2!
Pp
p
p+1
−k+1
+ (p + 1) · k=2
n
− k=3
n−k+2
k
k
.
P
p
n−k+1
(p + 1) · p + pk=2
k
Pp
Since each n-dependent term has a negative exponent, we will get in the limit as n → ∞,
(p + 1)p −
(p+1)!
(p−1)!2!
(p + 1)p
(p + 1)p −
=
Since
lim
(p+1)(p)(p−1)!
(p−1)!2!
(p + 1)p
=1−
1
1
= .
2
2
xn+1 − xn
1
= ,
yn+1 − yn
2
it follows from our theorem above that
p
n
xn
1
1 + 2 p + · · · + np
−
= lim
= .
lim
p
n
p+1
yn
2
(c) Want to compute
lim
(2(n + 1) + 1)p
(2n + 3)p
=
lim
.
(n + 1)p+1 − np+1
(n + 1)p+1 − np+1
Expanding,
(2n + 3)p
(n + 1)p+1 − np+1
Pp
p k
p k
p−k
p
3 (2n)−k
+ k=1
3 (2n)
2 + k=1
k
k
=
=
.
Pp+1 p + 1
Pp+1 p + 1
p
p−k
−k
(p + 1)n + k=2
(p + 1) + k=2
n
n
k
k
Pp
2p np
Taking the limit as n → ∞,
lim
By the theorem above,
lim
(2n + 3)p
2p
=
.
(n + 1)p+1 − np+1
p+1
1p + 3p · · · + (2n + 1)p
2p
=
.
np+1
p+1
2. Let (xn ) and (yn ) be two sequences defined recursively as follows: x1 = a ≥ 0, y1 = b ≥ 0,
xn+1 =
√
xn yn ,
yn+1 =
3
x n + yn
,
2
n≥1
Prove that the sequences (xn ) and (yn ) are convergent and that
lim xn = lim yn .
Solution:
Lemma. For any s, t ∈ R,
s2 + t2
.
2
st ≤
Proof. It is clear that
(s − t)2 ≥ 0
Expanding,
s2 − 2st + t2 ≥ 0 ⇒
In particular, if s =
√
u and t =
√
s2 + t2
≥ st
2
v, then for any u, v ≥ 0,
√
Then, for any n ≥ 1,
√
uv ≤
u+v
.
2
xn + yn
2
Since x1 , y1 ≥ 0, xn , yn ≥ 0 for n ≥ 1 by definition. We can therefore use our lemma to conclude
that for n ≥ 2, xn ≤ yn . Then for n ≥ 2,
p
√
xn+1 = xn yn ≥ x2n = xn ,
xn+1 =
xn yn ,
yn+1 =
xn + yn
2yn
≤
= yn .
2
2
Thus, (xn+1 ) is an increasing sequence and (yn+1 ) is a decreasing sequence. Since xn+1 ≤ yn+1 ≤
yn for n ≥ 3, we inductively conclude xn+1 ≤ y3 . Thus, (xn+1 ) is a monotonically increasing
sequence bounded above, and converges by the monotone convergence theorem. Since yn+1 ≥ 0
for n ≥ 0, (yn+1 ) is a monotonically decreasing sequence bounded below and converges by the
monotone convergence theorem. Letting L be the limit of (xn+1 ), and thus of (xn ), and noticing:
yn+1 =
yn =
we have:
lim yn =
3. Prove that
√
( xn yn )2
,
xn
(lim xn+1 )2
L2
=
= L = lim xn .
lim xn
L
1
1
1
< ln 1 +
<
n+1
n
n
for all n ∈ N.
Solution:
4
It was shown in class that
1
1+
n
n
1 n+1
<e< 1+
n
for all n ∈ N. Taking logarithms yields
"
#
1
1 n+1
1 n
1
= n ln 1 +
< 1 < ln 1 +
ln 1 +
= (n + 1) ln 1 +
n
n
n
n
1
1
The left part of this inequality implies that ln 1 +
< whereas the right part implies that
n
n
1
1
for all n ∈ N. Combining both inequalities yields
< ln 1 +
n+1
n
1
1
1
<
< ln 1 +
n+1
n
n
which is what we had to show.
4. Prove that the sequence
xn = 1 +
1
1 1
+ + · · · + − ln n,
2 3
n
n ∈ N,
converges.
Remark: The limit of this sequence is called the Euler-Mascheroni constant; its numerical value
is 0.5772156649.... It is currently unknown whether this constant is rational or irrational.
Solution:
We start by showing that (xn ) is decreasing; we will use the estimates from question 3.
1 1
1
1 1
1
1
xn − xn+1 = 1 + + + · · · +
− 1 + + + ··· + +
− ln n + ln(n + 1)
2 3
n
2 3
n n+1
1
n+1
1
1
1
− ln n + ln(n + 1) = −
+ ln
=−
+ ln 1 +
>0
=−
n+1
n+1
n
n+1
n
by question 4 i.e. (xn ) is decreasing. We will show next that (xn ) is bounded below by 0 i.e.
that xn ≥ 0 for all n ∈ N.
ln n = ln n − ln 1 = (ln n − ln(n − 1)) + (ln(n − 1) − ln(n − 2)) + · · · + (ln 2 − ln 1)
n
n−1
2
= ln
+ ln
+ · · · + ln
n−1
n−2
1
1
1
1
1
1
= ln 1 +
+ ln 1 +
+ · · · + ln 1 +
<
+
+ ··· + 1
n−1
n−2
1
n−1 n−2
by question 3. Thus
1 1
1
xn = 1 + + + · · · + − ln n >
2 3
n
1 1
1
1
1
1
1 + + + ··· +
−
+
+ ··· + 1 = > 0
2 3
n
n−1 n−2
n
Thus (xn ) is decreasing and bounded below and is thus convergent.
5
5. Prove that
lim
1
1
1
+
+ ··· +
n+1 n+2
2n
= ln 2.
Solution:
Let
1
1
1
+
+ ··· +
n+
2n 1 n+ 2 1
1
1
yn := ln 1 +
+ ln 1 +
+ · · · + ln 1 +
n
n+1
2n − 1
xn :=
and
zn :=
1
1
1
+
+ ··· +
n n+1
2n − 1
Then, by question 3, we have that xn < yn < zn for all n ∈ N.
For yn we obtain:
1
1
1
yn = ln 1 +
1+
... 1 +
n
n+1
2n
n+1 n+2 n+3
2n
= ln
·
·
···
n
n+1 n+2
2n − 1
(Note that this a telescoping product!)
= ln
Furthermore, zn = xn +
2n
= ln 2
n
1
1
1
−
= xn +
.
n 2n
2n
1
1
Combining these results yields xn < ln 2 < xn +
and thus ln 2 −
< xn < ln 2. Since
2n
2n
1
lim ln 2 −
= lim (ln 2) = ln 2, it follows from the squeeze theorem that (xn ) converges to
2n
ln 2.
6. Let
xn =
1+
1
2
1
1
1+
··· 1 + n ,
4
2
n ∈ N.
Prove that the sequence (xn ) converges.
Solution:
xn+1 =
1
1+
2
1
1
1
1
1+
··· 1 + n
1 + n+1 = xn 1 + n+1 > xn
4
2
2
2
6
The sequence (xn ) is thus increasing. We show next that (xn ) is bounded above by e, using
estimates obtained in question 3 as well as the formula for the sum of a finite geometric series
n−1
X
1 − rn
(recall that a + ar + ar2 + · · · + arn−1 =
).
a rk = a
1−r
k=0
1
1
1
ln(xn ) = ln 1 +
1+
··· 1 + n
2
4
2
1
1
1
= ln 1 +
+ ln 1 +
+ · · · + ln 1 + n
2
4
2
1
1
By question 3, ln 1 +
< . Thus
k
k
ln(xn ) <
1
1 1
1
1 1
1
1
+ + · · · + n = · 1 + · + · · · + · n−1
2 4
2
2
2 2
2 2
where the right-hand side is a finite geometric series with a =
1
2
and r = 12 . Thus
n
1
1 1 − 12
=1− n
ln(xn ) <
2 1 − 12
2
1 1
1
1
+ + · · · + n = 1 − n can also be proved by induction.)
2 4
2
2
Consequently ln(xn ) < 1 which means that xn < e for all n ∈ N.
(The formula
The sequence (xn ) is thus increasing and bounded above and thus converges.
7. Let (xn ), xn > 0 be a convergent sequence, then
lim
n→∞
√
n
x1 x2 x3 · · · xn = lim xn
n→∞
Solution:
Let L ≡ limn→∞ xn and let yn ≡
√
n
x1 x2 x3 · · · xn There are two possibilities.
First scenario is that L = 0. In this case, for a given ε > 0, there exists N 0 ∈ N such that for all
n > N 0,
0 < xn <
ε
2
Thus,
yn =
√
n
√
√
n
c n xN 0 +1 · · · xn
ε n−N 0
1
n
n
≤c ·
2
!1
0
2N c n ε
=
2
εN 0
1 ε
= (a) n
2
x1 · · · xN 0 · · · xn =
7
√
where c ≡ n x1 · · · xN 0 and a ≡
all n ≥ N 00 ,
0
2N c
0 .
εN
1
Since limn→∞ a n = 1, there exists N 00 ∈ N such that for
1
an < 2
Thus, choosing N > max {N 0 , N 00 }, we have
ε
yn < (2) = ε
2
for all n ≥ N . Thus, lim yn = 0.
In the second scenario, L > 0. In this case, by Theorem 3.2.3 (b) in Bartle and Sherbert, we
can equivalently show that
lim
n→∞
Thus, defining zn =
expression for zn :
yn
L,
yn
=1
L
our goal becomes to show that lim zn = 1. We now examine the
yn
xn
√
n x x ···x
1 2
n
=
L
r
x1 x2 · · · xn
= n
Ln
r x x1
x2
n
= n
···
L
L
L
zn =
Fixing ε > 0 (Without loss of generality, ε < 1) we may choose N 0 ∈ N large enough so that
|xn − L| < 2ε L for all n ≥ N 0 . In particular, we have xn > L − L 2ε for n ≥ N 0 . Thus,
r x 0 x1
x2
N
zn > n
···
L
L
L
1
ε
= cn 1 −
2
L 1−
L
ε
2
0
! n−N
n
where
c=
Q 0 xi !
2N 0 N
i=1 L
>0
(2 − ε)N 0
1
Once again, we shall use the fact that lim c n = 1 to choose N 00 large enough so that
1
cn > 1 −
ε
2
Thus, choosing N > max {N 0 , N 00 }, we have that
ε 2
ε2
zn > 1 −
=1−ε+
>1−ε
2
4
8
for n ≥ N .
In an analogous manner, one can ensure that N is suitably large enough so that for n ≥ N ,
zn < 1 + ε
(it is an exercise to show this rigorously). Summarizing, for a given ε > 0, we have found N ∈ N,
such that
−ε < zn − 1 < ε
for all n ≥ N . Equivalently stated,
|zn − 1| < ε ∀n ≥ N
This proves the claim.
8. Let (xn ), xn > 0 be a sequence such that the limit
xn+1
n→∞ xn
L = lim
exists. Prove that
lim
√
n
n→∞
xn = L.
Using this result, prove that
n
=e
lim √
n
n→∞
n!
Solution:
Let yn be a sequence defined by
y1 = x1
xn
yn =
for n ≥ 2
xn−1
Then, one can easily verify that yn > 0 for all n ∈ N and that
lim yn = L
n→∞
We may thus use the result from Exercise 7 to conclude that
lim
n→∞
√
n
y1 y2 · · · yn = L
However, we see that
y1 y2 · · · yn = x1
x2
x1
9
x3
x2
···
xn
xn−1
= xn
and thus
lim
√
n
n→∞
xn = L.
To prove the second statement in the problem, we define the sequence (xn ) by
xn ≡
nn
.
n!
We then examine the ratio of consecutive terms in the sequence,
(n + 1)n+1 n!
xn+1
= n
xn
n (n + 1)!
(n + 1)(n + 1)n n!
=
(n + 1)n!nn
n+1 n
=
n
1 n
= 1+
n
Thus, by our definition of the number e, we have
xn+1
=e
n→∞ n
lim
which, using the first result of this problem, implies that
lim
√
n
n→∞
n
xn = √
= e.
n
n!
9. Let (xn )∞
n=1 be a positive subadditive sequence. That is, for any n, m ∈ N, we have
0 ≤ xn+m ≤ xn + xm
Show that limn→∞
xn
n
exists and
nx
o
xn
n
= inf
:n∈N
n→∞ n
n
lim
Solution:
We let L ≡ inf
x
n
n
: n ∈ N . We must clearly have
L≤
xn
n
∀n ∈ N
To show convergence, we let ε > 0 be given and choose k ∈ N, such that
ak
ε
<L+
k
2
10
(1)
Note that this is always possible by the definition of L. Now for each n ≥ k, we have n = pn ·k+rn
for some pn ∈ N and rn ∈ {0, . . . , k − 1}. By inductively applying the subadditivity condition,
we find that
xn = xpn ·k+rn ≤ xpn ·k + xrn ≤ pn xk + xrn
dividing by n, we find that
xn
pn xk
xr
xk
xr
ε xr
≤
+ n ≤
+ n <L+ + n
n
pn · k + rk
n
k
n
2
n
(2)
for all n ≥ k.
Now, since rn ∈ {0, . . . , k − 1}, we have that
xrn ≤ max {x1 , . . . , xk−1 } ≡ M
, equations (??) and (??) imply that for all
Thus, choosing N ∈ N such that N > max k, 2M
ε
n ≥ N , we have
L≤
xn
≤L+ε
n
and since ε was an arbitrary positive number, we have completed the proof.
10. Let (xn ) be a bounded sequence and for each n ∈ N let sn = sup{xk : k ≥ n} and S = inf{sn }.
Show that there exists a subsequence of (xn ) that converges to S.
Solution:
First note that for n ∈ N
sn+1 = sup{xk : k ≥ n + 1} = sup{xn+1 , xn , xn−1 , . . . }
= sup({xn+1 } ∪ {xn , xn−1 , . . . })
= max{xn+1 , sup{xn , xn−1 , . . . }}
≤ sup{xn , xn−1 , . . . } = sn .
Since (xn ) is bounded, ∃M ∈ R such that |xn | ≤ M for all n ∈ N. By definition, we have
that sn ≥ xn ≥ −M . Therefore, since n ∈ N was arbitrary, (sn )n∈N is bounded below by −M
and decreasing and we conclude that S := inf sn = limn→∞ sn . We will inductively construct a
subsequence (xnk )k∈N which converges to S.
The first element Simply take xn1 = x1 .
The k -th element (k > 1) Assume we constructed xnk−1 . We want to find xnk such that
|xnk − S| < 1/k. Since, S = limm→∞ sm , ∃Mk0 ∈ N such that |sm − S| < 1/(2k) for any
m ≥ Mk0 . In particular, we can find Mk > nk−1 such that |sMk − S| < 1/(2k). Now since
sMk = supn≥Mk xn , ∃nk ≥ Mk > nk−1 such that sMk ≤ xnk < sMk + 1/(2k). Then,
|xnk − S| ≤ |xnk − sMk | + |sMk − S| < 1/(2k) + 1/(2k) = 1/k.
Since the sequence (xnk )k∈N satisfies |xnk − S| < 1/k for all k, by the Archimedean property,
∀ > 0, ∃K() ∈ N such that 1/K() < ⇒ |xnk − S| < ∀k ≥ K()
and we conclude xnk → S as k → ∞.
11
11. Let L ⊂ R. The set L is said to be open if for any x ∈ L there exists > 0 such that
(x − , x + ) ⊂ L. The set L is said to be closed if its complement Lc = {x ∈ R : x ∈
/ L} is open.
(a) Prove that L is closed if and only if for any convergent sequence (xn ) with xn ∈ L, the
limit x = limn→∞ xn = x is also an element of L.
(b) Let (xn ) be a bounded sequence. A point x ∈ R is called an accumulation point of (xn ) if
there exists a subsequence (xnk ) of (xn ) such that limk→∞ xnk = x. We denote by L the
set of all accumulation points of (xn ). By the Bolzano-Weierstraß Theorem, the set L is
non-empty. Prove that L is a bounded closed set.
(c) Let (xn ) be a bounded sequence, let L be as in part (b) and let S be as in problem 1. Prove
that S = sup L.
Solution:
(a)(⇒) Let L be a closed set. Let (xn )n∈N be converging sequence with xn ∈ L and limn→∞ xn =
x. Then, in particular, ∀ > 0, ∃n ∈ N such that |xn − x| < . Therefore, ∀ > 0,
can find xn ∈ L such that xn ∈ (x − , x + ). Therefore, ∀ > 0, (x − , x + ) 6⊆ Lc .
Equivalently, @ > 0 such that (x − , x + ) ⊆ Lc . Since Lc is open, we must conclude
that x ∈
/ Lc and therefore that x ∈ L.
(⇐) Let L be a set that is not closed. Then Lc is not open, which implies that ∃x ∈ Lc such
that ∀n ∈ N, (x − n−1 , x + n−1 ) 6⊆ Lc ⇔ (x − n−1 , x + n−1 ) ∩ L 6= ∅. Hence, ∀n ∈ N,
∃xn ∈ L such that xn ∈ (x − n−1 , x + n−1 ). We have constructed a sequence (xn )n∈N
in L with the property that ∀n ∈ N, |xn − x| < n−1 , which implies that limn→∞ xn = x
where x ∈
/ L. Therefore, it is not true that for any converging sequence (xn )n∈N with
xn ∈ L, the limit x = limn→∞ xn is also an element of L.
Remark: A lot of people did well on the first implication. For the second, a lot of people have confused
“L is not closed” and “L is open”: they do not mean the same thing. Some sets are both open and
closed (clopen) and some sets are neither closed nor open.
Also note that closed [resp. open] sets are not necessarily closed [resp. open] intervals and that
in general, (x − , x + ) 6⊂ Lc does not imply (x − , x + ) ⊂ L.
(b) Let ` ∈ L be an accumulation point. Then there is a subsequence (xnk )k∈N such that
limk→∞ xnk = `. In particular, ∃K ∈ N such that |xnK − `| < 1. This implies (write the
details) |`| < 1 + |xnK |. Since (xn )n∈N is a bounded sequence, ∃M ∈ R such that M ≥ |xn |
for all n ∈ N and in particular, M ≥ |xnK |. Therefore, |`| < 1 + M . Since ` ∈ L was
arbitrary, we conclude L is bounded (by M + 1).
Let (`m )m∈N be a convergent sequence with `m ∈ L and ` = limm→∞ `m . We want to
construct a subsequence (xnM (j),K(M (j)) )j∈N of (xn )n∈N that converges to `. Let xnM (1),K(M (1)) =
x1 . For j > 1, since ` = limm→∞ `m , ∃M (j) ∈ N such that |`M (j) − `| < 1/(2j). Since
`M (j) ∈ L, we can find a subsequence (xnM (j),k )k∈N such that limk→∞ xnM (j),k = `M (j) . In
particular, ∃K(M (j)) ∈ N such that |xnM (j),K(M (j)) − `M (j) | < 1/(2j) and nM (j),K(M (j)) >
nM (j−1),K(M (j−1)) . This defines a subsequence (xnM (j),K(M (j)) )j∈N of (xn )n∈N satisfying
xnM (j),K(M (j)) − ` ≤ xnM (j),K(M (j)) − `M (j) + `M (j) − `
< 1/(2j) + 1/(2j)
= 1/j.
12
Using the Archimedean property as in part (a), we get that limj→∞ xnM (j),K(M (j)) = ` and
we conclude ` ∈ L.
Remark: One has to be careful with notation in order not to give new meaning to objects that were already
defined. This might require a lot of indices.
(c) We first need to show that S is an upper bound of L. Let ` ∈ L. Then, there is a
subsequence (xnk )k∈N such that limk→∞ xnk = `. Since nk ≥ k, we have xnk ≤ sk , which
implies that limk→∞ xnk ≤ limk→∞ sk , that is ` ≤ S (using Problem 1). Since ` ∈ L was
arbitrary, we conclude L is bounded above by S and S ≥ sup L.
By Problem 1, S is an accumulation point of (xn ), i.e. S ∈ L, so that S ≤ sup L. We
conclude S = sup L.
12. Using the Cauchy Convergence Criterion, prove that the sequence
xn = 1 +
1
1
+ ··· + 2
2
2
n
is convergent.
Solution:
Let > 0. By the Archimedean property, ∃N ∈ N such that N > −1 . Let n, m ∈ N satisfy
n ≥ N and m ≥ N . Without loss of generality, m > n. Then,
1
1
1
+
+ ··· + 2
(n + 1)2 (n + 2)2
m
1
1
1
<
+
+ ··· +
n(n + 1) (n + 1)(n + 2)
(m − 1)m
1
1
1
1
1
1
−
+
−
+ ··· +
−
=
n n+1
n+1 n+2
m−1 m
1
1
= −
n m
1
< .
<
N
|xm − xn | =
Since > 0 was arbitrary, we conclude (xn ) is a Cauchy sequence and therefore convergent.
13
13. Definition: a sequence (xn ) has bounded variation if there exists c > 0 such that for all n ∈ N,
|x2 − x1 | + |x2 − x2 | + · · · + |xn − xn−1 | < c.
Show that if a sequence has bounded variation, then the sequence is converging. Find an example
of a convergent sequence which does not have bounded variation.
Solution:
We show the contrapositive: if a sequence is not converging, then it does not have bounded
variation. Assume (xn ) does not converge. Then, (xn ) is not Cauchy, i.e.
∃0 such that ∀N ∈ N, ∃n(N ), m(N ) ≥ N such that |xm(N ) − xn(N ) | ≥ 0 .
(3)
Let c > 0 be arbitrary. Then, by the Archimedean property, ∃K ∈ N such that K0 ≥ c.
1. By (??) with N = 1, there exists n(1), m(1) ∈ N such that |xm(1) − xn(1) | ≥ 0 without loss
of genreality m(1) > n(1).
2. By (??) with N = m(1) + 1, there exists m(2) > n(2) ≥ m(1) + 1 > m(1) > n(1) such that
|xm(2) − xn(2) | ≥ 0 .
..
.
K. By (??) with N = m(K − 1) + 1, we get m(K) > m(K) > m(K − 1) > n(K − 1) such that
|xm(K) − xn(K) | ≥ 0 .
Then,
c ≤ K0
≤ |xm(1) − xn(1) | + |xm(2) − xn(2) | + · · · + |xm(K) − xn(K) |
≤ |x2 − x1 | + |x2 − x3 | + · · · |xm(K)−1 − xm(K) |.
(Write the details). Since c > 0 was arbitrary, we conclude (xn ) does not have bounded variation.
Consider the sequence xn = (−1)n /n. It is easy to show, using the Archimedean property, that
(xn ) converges to 0. Note that (write the details)
|x2 − x1 | + |x3 − x2 | + · · · + |xn − xn−1 | ≥ 1 +
1
1
+ ··· +
2
n
and that there exists no c > 0 such that 1 + 21 + · · · + n1 < c for all n ∈ N (see Example 3.3.3(b)
in Bartle and Sherbert, fourth edition). We conclude (xn ) does not have bounded variation.
14
14. Let x1 < x2 be arbitrary real numbers and
xn =
xn−1 2xn−2
+
,
3
3
n > 2.
Find the formula for xn and lim xn .
Solution:
The characteristic equation is λ2 = λ/3 + 2/3, which has roots λ1 = −2/3 and λ2 = 1. The nth
term is therefore given by
xn = (−2/3)n c1 + 1n c2 .
We have x1 = (−2/3)c1 + c2 and x2 = (4/9)c1 + c2 . Solving the linear system (write the details)
yields c1 = −9(x1 − x2 )/10 and c2 = (2x1 + 3x2 )/5, so that
xn = (−2/3)n (−9(x1 − x2 )/10) + (2x1 + 3x2 )/5.
Since (−2/3)n → 0 as n → ∞, we have (−2/3)n (−9(x1 − x2 )/10) → 0 as n → ∞ and xn =
(−2/3)n (−9(x1 − x2 )/10) + (2x1 + 3x2 )/5 → (2x1 + 3x2 )/5 as n → ∞.
15
15. Let x1 > 0 and
xn+1 =
1
,
2 + xn
n ≥ 1.
Show that (xn ) is a contractive sequence and find lim xn .
Solution:
We first show by induction that xn > 0 for all n ∈ N. Base case n = 1 is given. Assume xn > 0.
1
Then, 2 + xn > 0 so that xn+1 = 2+x
> 0.
n
Now, we have
|xn+2 − xn+1 | = 1 2 + xn − 2 − xn+1 1
−
=
2 + xn+1 2 + xn (2 + xn+1 )(2 + xn ) |xn − xn+1 |
|xn − xn+1 |
=
<
|(2 + xn+1 )(2 + xn )|
4
since xn+1 > 0 and xn > 0. This shows (xn ) is contractive. Therefore, it is Cauchy and it
converges. Let x = lim xn . Then,
2 + lim xn = 2 + x
⇒
1
1
1
=
=
2 + xn
2 + lim xn
2+x
1
1
x = lim xn+1 = lim
=
2 + xn
2+x
2
x + 2x = 1
⇒
x = −1 ±
⇒
lim
⇒
√
2.
√
However, since xn > 0 for all n ∈ N, we must have lim xn ≥ 0 and we conclude lim xn = −1+ 2.
16