sequence

2(a)

un+1=√(3un)

u1=1,u2=√3

So when n=1, the statement is true

Assume that uk+1>=uk

√uk+1>=√uk
So √(3uk+1)>=√(3uk)
So uk+2>=uk+1

3(a)
for n=1
a1=1/2<1/√3

Assume that ak<1/√(2k+1)
ak+1
=ak*(2k+1)/(2k+2)
<√(2k+1)/(2k+2)
<√(2k+2)/(2k+2)
=1/√(2k+2)

(b)
b1=3/2>√2
n=1 is true
Assume bk is true
bk+1
=bk*(2k+3)/(2k+2)
>√(k+1)*(2k+3)/(2k+2)
=(1/2)[(2k+3)/√(k+1)]
>√(k+2)]

We want to prove that (2k+3)^2>4(k+1)(k+2)
Since LHS=4k^2+12k+9>4k^2+12k+8=RHS

(d)(iii)
an<1/√(2n+2)
bn>√(n+1)

cn^2=anbn

lim an=1/√2
lim bn=infinity

Now notice that bn>an
So cn^2>an^2
cn>an
and lim cn>lim an=1/√2

For the other inequality
Since an < 1/√(2n+1)
And cn=an*√(2n+1)
For each of cn, it should be lower than
1/√(2n+1)*√(2n+1)=1

So lim cn<1

2008-04-06 17:17:32 補充：
&lt <
&gt >

仍然是亂碼﹐都無改善

2008-04-06 20:23:25 補充：
Assume that ak<1/√(2k+1)
ak+1
=ak*(2k+1)/(2k+2)
<√(2k+1)/(2k+2)
<1/√(2k+3)

We want to prove that (2k+1)(2k+3)<(2k+2)^2
Since LHS=4k^2+8k+3<4k^2+8k+4=RHS