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.

(c) Proof: Suppose ( s_{n}^{)} → A and ( s_{n}^{+} t_{n}^{)} → B . Then

t_{n}^{= (} s_{n}^{+} t_{n}^{)} - s_{n}^{→} B - A

(d) Counterexample: ( s_{n}^{) =}

parenleftbigg

1 ,

1

2

,

1

3

, · · · ,

1

n

· · ·

parenrightbigg

, s_{n}^{→} 0, ( t_{n}^{) = ((} -

1)^{n}_{) =}

( - 1 , 1 , - 1 , 1 , . . . ) does not converge. But ( s_{n}^{t}_{n}^{) =}

parenleftbigg

( -

1)^{n}

n

parenrightbigg

→

0.

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

parenleftbigg

1

n

parenrightbigg

, ( s_{n}^{) =}

parenleftbigg

n

n_{2}^{+ 1}

parenrightbigg

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

n^{2}

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 .

1

| s^{2}

n^{-} s^{2}^{|} = | 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

s^{2}

n^{-} s^{2} 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 s^{2}

n^{→} s^{2}_{.}

15. (a)

√

n + 1 -

√

n =

parenleftBig √

n + 1 -

√

n

parenrightBig √

n + 1 +

√

n

√

n + 1 +

√

n

=

1

√

n + 1 +

√

n

<

1

2

√

n

→

0

(b)

√

n_{2}^{+ 1} - n =

parenleftBig √

n_{2}^{+ 1} - n

parenrightBig

√

n^{2}^{+ 1 +} n

√

n_{2}^{+ 1 +} n

=

1

√

n_{2}^{+ 1 +} n

<

1

2 n

→ 0

(c)

√

n^{2}^{+} n - n =

parenleftBig √

n^{2}^{+} n - n

parenrightBig

√

n_{2}^{+} n + n

√

n_{2}^{+} n + n

=

n

√

n_{2}^{+} n + n

→

1

2

17. Let s_{n}^{=} k^{n}^{/n} !.

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