Summary lectures solution - lecture problem

lOMoARcPSD|16248954




M A 101 ( M a t h e m a t i c s I )
H i n t s / E x p l a n a t i o n s for E x a m p l e s i n Lectures


Sequence

E xa m p le : The sequence ( 2nn++13 ) is convergent with2 limit 1 .
P roof: L et ε > 0. For all n ∈ N, we have | 2nn++13 —12| = 14n + 6 < 41n . T here exists n 0 ∈ N such that
n 0 > 14ε . Hence | n2 n+ +1 3 — 12 | < 14n0 < ε for all n ≥ n 0 and so the given sequence is convergent with
limit 21 .

E x am ple: T he sequence (1, 2, 1, 2, ...) is not convergent.
Proof: If possible, let the given sequence ( x n ) (say) be convergent with limit l . Then there exists
n 0 ∈ N such that |x n − l | < 12 for all n ≥ n 0. Hence |x 2n 0 —l | < 12 and |x 2 n 0+ 1 − l | < 12 and so
|2 − l | < 12 and |1 − l | < 12 . This gives 1 = |(2 − l ) −(1 − l )| ≤ |2 − l | + |1 − l | < 1, which is a
contradiction. Therefore the given sequence is not convergent.

E xa m p le : The sequence ( n 3 + 1) is not convergent.
Proof: If possible, let ( n 3 + 1) be convergent. Then there exist l ∈ R and n 0 ∈ N such that
|n3 + 1 − l | < 1 for all n ≥ n 0 ⇒ n 3< l for all n ≥ n 0 , which is not true. Therefore the given
sequence is not convergent.

E x am ple: T he sequence ( 32nn++ 52 ) is bounded.
P roof: For all n ∈ N, |32nn ++ 25 | = 32nn ++ 25 < 3 n2n+ 2 = 3
2
+ 1
n
≤ 25 . Hence the given sequence is bounded.

E x am ple: T he sequence (1, 2, 1, 3, 1, 4, ...) is unbounded.
Proof: If possible, let the given sequence ( x n ) (say) be bounded. Then there exists M > 0 such
that |xn| ≤ M for all n ∈ N. This gives n ≤ M for all n ∈ N, which is not true. Therefore the
given sequence is unbounded.
2
E xa m p le : The sequence ( 3 2nn2 +−53nn+ 3 ) is convergent with limit 23.
2− n3
Proof: We have 2 n 2− 3 n
3 n2 + 5 n + 3
= 3 + n5 + n32
for all n ∈ N. Since 1
n
→ 0, the limit rules for algebraic
2− 0 = 2 .
operations on sequences imply that the given sequence is convergent with limit 3+ 0+ 0 3
√ √
E xa m p le : The sequence ( n + 1 − n ) is convergent with limit 0.
√ √ √1
Proof: For all n ∈ N, n + 1 − n = √ n + 11 + √ n = √ n1 . Since n1 → 0, the limit rules for al-
1+ n +1
0
gebraic operations on sequences imply that the given sequence is convergent with limit √
1+ 0+ 1
= 0.

E xa m p le : If |α| < 1, then the sequence ( α n ) converges to 0.
Proof: If α = 0, then α n = 0 for all n ∈ N and so ( α n ) converges to 0. Now we assume that
1
α / = 0. Since |α| < 1, |α1 | > 1 and so |α | = 1 + h for some h > 0. For all n ∈ N, we have
(1 + h ) n = 1 + nh + n ( n2!− 1) h 2 + ···+ hn > nh ⇒ |α |n = ( 1 +1h ) n < nh
1 for all n ∈ N. G iven ε > 0, we

choose n 0 ∈ N satisfying n 0 > hε 1 . Then |αn −0| = |α| n < 1 < ε for all n ≥ n and hence ( α n)
n0h 0
converges to 0.
Alternative proof: Given ε > 0, we choose n 0 ∈ N satisfying n 0 > log log ε
|α |
. Then for all n ≥ n 0, we
have |α n −0| = |α| n ≤ |α| n 0 < ε and hence ( α n ) converges to 0. (This proof assumes the definition
of logarithm.)

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1
E xa m p le : If α > 0, then the sequence ( α n ) converges to 1.
Proof: We first assume that α ≥ 1 and let x n = α n − 1 for all n ∈ N. Then x n ≥ 0 and
1


α = (1 + x n) n = 1 + nx n + n ( n2!− 1) 2 nfor all n ∈n αN.x +n ···+ x > nx for all nn ∈ N.
n So 0 ≤ x <
n
→
α 1. If α < 1, then→ 0, by sandwich theorem, it follows that x → 0. n Consequently α Since
1
n n
→ 1.→ 1. It follows that 1 1 1
α α
αn > 1 and as proved above,n ( )
Alternative proof: We first assume that α ≥ 1. For each n ∈ N, applying the A.M. ≥ G.M.
. Since ≤ 1 + inequality for the numbers 1, ..., 1, α (1 is repeated n −1 times), −1 1 ≤ α
1
n weαget
n
α − 1 → 0, by sandwich theorem, it follows that α n → 1. The case for α < 1 is same as given in
1
n
the above proof.
1
Examp le: The sequence ( n n ) converges to 1.
−1. T hen for all n ∈ N, nn = (1 + a ) n = 1 + naP roof: For all n ∈ N, nlet a = n n + n ( n2!− 1) an2 +
1
n

···+ ann > n ( n − 1)na− 1n ⇒2 0 ≤2 an < 2!2 for all n ∈ N. Since n −21 → 0, by sandwich theorem, it follows
→ 1.n2 that a → 0 and n so a → 0. C onsequently
1
n n
1
E xa m p le : The sequence ((2n + 3n ) n ) converges to 3.
< 2 .3 for all n n∈ N. n Also,P roof: We have 3 n< 2 + n3 < n 2.3 n forn all nn ∈ N ⇒ 3 < (2 + 3 )
1 1


both the sequences (3 , 3, ...) and (2 n .3) converge to 3. (Note that 2 n → 1.) Hence by sandwich
1 1


theorem, the given sequence converges to 3.
≤ 3[1+( ) ]for all n ∈ N, we
n n 1 1 1
n
= 3[1+( ) ]Alternative proof: Since (2 + 323 n) n n have 2 n)
n n 3 < (2 + 3
3
2n
for all n ∈ N. Also, both the sequences (3 , 3, ...) and (3[1 + ( 3 ) ]) converge to 3. (Note that
(32 )n → 0.) Hence by sandwich theorem, the given sequence converges to 3.

E xa m p le : The sequence ( √ n 2 + 11 + ···+ √ n 2 + n1 ) converges to 1.
Proof: We have √ ≤√ 1
n2+n
√ + ···+
n2+1 n2+n
n1 ≤ √ n
n2+ 1
for all n ∈ N. Also, √ n
n2+ n
= √1 1
→1
1+ n
and √ n
n2+ 1
= ,
1
1 → 1. Hence by sandwich theorem, the given sequence converges to 1.
1+
n2

) is convergent.E xa mp le : If α ∈ R, then
n
α the sequence (
n!
Proof: Let x n = αn! for all n ∈ N. If α = 0, then x n = 0 for all n ∈ N and so ( x n ) converges to 0.
n


If α / = 0, then lim x n + 1| = lim|n + 1|xα| = 0 < 1 and so ( xn ) converges to 0.
n→ ∞ n→∞ n

n
E xa m p le : The sequence ( 2n 4 ) is not convergent.
2
n 2 > 1. Therefore the sequence|( 1=+ )lim forx1nall
4 n ∈ N, then lim |Proof : If x
n xn+1 2
n4 = n→ ∞ n→ ∞
=
n
( x n ) is not convergent.
1 sequence (1 −
) is increasing.E xa mp le : The
n
P roof: For all n ∈ N, n + 11 n 1 for all nn +∈11 N. T herefore
n
1 the given sequence> 1 − and so 1 −<
is increasing.

E xa m p le : The sequence ( n + n1 ) is increasing.
) = 1 −) −( n + Proof: For1n all n ∈ N, 1 (n + 1 +
n+ 1
1
n ( n + 1)
> 0 ⇒ n + 1+ n+ 1
1
> n+ n
1 for all
n ∈ N. Therefore the given sequence is increasing.

E xa m p le : The sequence (cos nπ
3
) is not monotonic.
6π 3π
and hence
3
3π 1 π
< cos 3 > cos 3 6π
= 1, we have 3cos= −1 and
π
cos,3 cos= 2Proof
3
: Since cos
the given sequence is neither increasing nor decreasing. Consequently the given sequence is not
monotonic.

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E xa m p le : Let x 1 = 1 and x n + 1 = 1
3
(xn + 1) for all n ∈ N. Then the sequence ( x n ) is con-
vergent and lim x n = 12.
n→ ∞ 1
Proof: For all n ∈ N, we have x n + 1 — x n = 13 (1 —2 x n ). Also, x 1 > 2 and if we assume that
1 1 1
x k > 12 for some k ∈ N, then x k + 1 = 13 ( x k + 1) > 3 ( 2 + 1) = 2 . Hence by the principle of
mathematical induction, x n > 12 for all n ∈ N. So ( x n ) is bounded below. Again, from above, we
get x n + 1 − x n < 0 for all n ∈ N ⇒ x n + 1 < x n for all n ∈ N ⇒ ( x n ) is decreasing. Therefore ( x n )
is convergent. L et l = lim x n. T hen lim x n + 1 = l and since x n + 1 = 13 ( x n + 1) for all n ∈ N, we
n→ ∞ n→ ∞
get l = 1( l
3
+ 1) ⇒ l = 1.
2

Alternative proof for showing that ( x n ) is decreasing: We have x 2 = 2 3< 1 = x 1 and if we assume
that x k + 1 < x k for some k ∈ N, then x k + 2 = 13 ( x k + 1 + 1) < 13 ( x k + 1) = x k + 1 . Hence by the
principle of mathematical induction, x n + 1 < x n for all n ∈ N.

E xa m p le : Let x n = 1 + 1!1 + 2!1 + ···+ n1! for all n ∈ N. Then the sequence ( x n ) is conver-
gent.
Proof: For all m, n ∈ N with m > n , we have | x m − x n | = ( n +1 1)! + ( n +1 2)! + ···+ m1 ! ≤
2 1 2 2
1
2n
+ 2 n1+ 1 + ···+ 2 m1−1 = 2 n (1 − 2 m − n ) < 2 n < n . Given ε > 0, we choose n 0 ∈ N satisfying
n 0 > 2ε . Then for all m, n ≥ n 0, we get |x m − x n| < n20 < ε. Consequently ( x n ) is a Cauchy
sequence in R and hence ( x n ) is convergent.

E xa m p le : Let 0 < α < 1 and let the sequence ( x n ) satisfy the condition | x n +1 − x n | ≤ α n
for all n ∈ N. Then ( x n ) is a Cauchy sequence.
Proof: For all m, n ∈ N with m > n , we have | x m − x n | ≤ |xn − x n + 1 | + | x n +1 − x n + 2 | + ···+
|xm − 1 − x m| ≤ α n + α n + 1 + ···+ α m − 1 = 1−α α (1 − α m − n ) < 1−α α . Since 0 < α < 1, α n → 0 and
n n


so given any ε > 0, we can choose n 0 ∈ N such that 1− α n 0 < ε. Hence for all m, n ≥ n , we have
α 0
| x m − x n | < 1α− α0 < ε. Therefore ( x n ) is a Cauchy sequence.
n




E xa m p le : Let 0 < α < 1 and let the sequence ( x n ) satisfy the condition | x n +2 − x n + 1 | ≤
α | x n +1 − x n | for all n ∈ N. Then ( x n ) is a Cauchy sequence.
Solution : For all m, n ∈ N with m > n , we have | x m − x n | ≤ |xn − x n + 1 | + | x n +1 − x n + 2 | + ···+
|x m − 1 −x m| ≤ ( α n − 1 + α n + ···+ α m − 2 )|x 2 −x 1| = α1− α (1 −α m − n )|x 2 −x 1| ≤ α1− α |x 2 −x 1|. Since
n −1 n −1


0 < α < 1, α n − 1 → 0 and so given any ε > 0, we can choose n 0 ∈ N such that α1−0 α |x 2 − x 1 | <
n −1


Hence for all m, n ≥ n 0, we have |x m − x n| ≤ α1−0 α |x 2 − x 1 | < ε. T herefore ( x n )ε.is a C auchy
n −1


sequence.

E x am ple: L et x 1 = 1 and let x n + 1 = 1 for all n ∈N. Then the sequence ( x n ) is conver-
√ xn+2
gent and lim x n = 2 −1.
n→ ∞
P roof: For all n ∈ N, we have |x n + 2 −x n + 1 | = |x n +11 + 2 —x1 n + 2 | = || xx n + 1n++21 |−| xx nn+| 2 | . Now, x 1 > 0 and if we
assume that x k > 0 for some k ∈ N, then x k + 1 = x 1+ 2 > 0. Hence by the principle of mathematical
k
induction, x n > 0 for all n ∈ N. Using this, we get |x n + 2 —x n + 1 | ≤ 14 |x n + 1 —x n | for all n ∈ N. It
follows that ( x n ) is a Cauchy sequence in R and hence ( x n ) is convergent. Let l = lim x n . Then
√ n→ ∞
lim x n + 1 = l and since x n + 1 = for all n ∈ N, we get l =
1 1 ⇒ l = −1 ± 2. Since x n > 0
n→ ∞ √ xn+2 l+ 2
for all n ∈ N, we have l ≥ 0 and so l = 2 −1.

E xa m p le : If x n = (−1) n(1 − n1 ) for all n ∈ N, then x n /→ 1. In fact, ( x n ) is not conver-
gent.
Proof: Since x 2n − 1 = (−1) 2 n − 1 (1 − 21n − 1 ) = 2n1− 1 − 1 → −1, x n /→ 1. Again, since x 2 n =
(−1) 2n(1 − 2 n1) = 1 − 2 n 1→ 1 / = −1, ( x n ) is not convergent.