Math 5333 Solns-Assignment4 - Solutions Homework#4 Section 4.2 6(a Counterexample(sn =(1 1 1 1(tn =(1 1 1 1(b Counterexample the example in part(a will

Solutions, Homework #4

Section 4.2

6. (a) Counterexample: ( s_n^{) = (1} , - 1 , 1 , - 1 , · · · ); ( t_n^{) = (} - 1 , 1 , - 1 , 1 , · · · ).

(b) Counterexample: the example in part (a) will work here, too.

t_n^{= (} s_n⁺ t_n⁾ - s_n^→ B - A

(d) Counterexample: ( s_n^{) =}

parenleftbigg

1 ,

, · · · ,

· · ·

parenrightbigg

, s_n^→ 0, ( t_n^{) = ((} -

1)ⁿ_{) =}

( - 1 , 1 , - 1 , 1 , . . . ) does not converge. But ( s_n^t_n^{) =}

parenleftbigg

( -

1)ⁿ

parenrightbigg

→

8. (a) There are lots of examples. Here are two: ( s_n^{) =}

parenleftbigg

parenrightbigg

, ( s_n^{) =}

parenleftbigg

n₂^{+ 1}

parenrightbigg

(b) There are lots of examples. Here are two: ( s_n^{) = (} n ) = (1 , 2 , 3 , 4 , . . . ) ( s_n^{) =} parenleftbigg

n²

n + 1

parenrightbigg

9. ( s_n⁾ and ( t_n ) are sequences such that s_n^≤ t_n^{for all} n .

(a) Choose any number M ∈ R . Since lim

n →∞

s_n^{= +} ∞ , there is a positive

integer N such that s_n^{> M} for all n > N . Since t_n^≥ s_n^{for all} n , it

follows that t_n^{> M} for all n > N . Therefore lim_n →∞^t_n^{= +} ∞ :

(b) Choose any number M ∈ R . Since lim_n →∞^t_n⁼ -∞ , there is a positive

integer N such that t_n^{< M} for all n > N . Since s_n^≤ t_n^{for all} n , it

follows that s_n^{< M} for all n > N . Therefore lim_n →∞^s_n⁼ -∞ :

12. Let epsilon1 >

0. Since ( s_n ) is convergent, it is bounded; | s_n^{| ≤} M for all n .

| s²

n^- s²^| = | s_n^- s | | s_n⁺ s | ≤ | s_n^- s | ( M + | s | ). Since s_n^→ s there is a

positive integer N such that | s_n^- s | <

epsilon1

M + | s |

for all n > N . Therefore

vextendsingle vextendsingle

s²

n^- s² vextendsingle vextendsingle = | s_n^- s | | s_n⁺ s | ≤ | s_n^- s | ( M + | s | ) < ( M + | s | )

epsilon1

M + | s |

= epsilon1

for all n > N and it follows that s²

n^→ s²_.

15. (a)

√

n + 1 -

√

n =

parenleftBig √

n + 1 -

√

parenrightBig √

n + 1 +

√

n + 1 +

√

n + 1 +

√

→

(b)

√

n₂^{+ 1} - n =

parenleftBig √

n₂^{+ 1} - n

parenrightBig

√

n²^{+ 1 +} n

√

n₂^{+ 1 +} n

√

n₂^{+ 1 +} n

2 n

→ 0

(c)

√

n²⁺ n - n =

parenleftBig √

n²⁺ n - n

parenrightBig

√

n₂⁺ n + n

√

n₂⁺ n + n

√

n₂⁺ n + n

→

17. Let s_n⁼ kⁿ^/n !.

Solutions, Homework #4 Section 4.2 8.(a) There are lots of examples.Here are two:(sn) =parenleftbigg1nparenrightbigg,(sn) =parenleftbiggnn2+ 1parenrightbigg(b) There are lots of examples. Here are two: (sn) = (n) = (1,2,3,4, . . .)(sn) =parenleftbiggn2n+ 1parenrightbigg9. (sn)and (tn) are sequences such thatsn≤tnfor alln.(a) Choose any numberM∈R. Since (b) Choose any numberM∈R. Since 1

You've reached the end of your free preview.

Want to read all 7 pages?

TERMFall '08
PROFESSOR Staff
TAGS Math, lim, Natural number, Sn

Share this link with a friend:

Copied!

Math 5333 Solns-Assignment4 - Solutions Homework#4 Section...

You've reached the end of your free preview.

Share this link with a friend:

Other Related Materials