Double Inequality: Find n0, c1, c2

[atrus_ovis]

Homework Statement

find n₀,c₁,c₂ for which the following is true:
c₁ n^b <=(n-a)^b<=c₂(n-a)^b , for all n > n₀

Homework Equations

http://en.wikipedia.org/wiki/Binomial_theorem" ?

The Attempt at a Solution

c₁ n^b <=(n-a)^b<=c₂(n-a)^b
c₁ n^b <=n^b-n^b-1a+n^b-2a²-...-a^b<=c₂n^b

c₁<=1-a/n + a²/n²-... -a^b/n^b<=c₂

isn't this true, for c₁=0 , c₂=1 , n₀ >a ? Since every fraction will be less than its predecessor in the alternating series

0<=1-a/n + a²/n²-... -a^b/n^b<=1

 

[micromass]

Yes, that seems to be correct.

 

[atrus_ovis]

Okay , may i ask another one then?
Prove if
f(n) <= (g(n)-f(n) ) *c/2 (1) , for constant c, for all n >= n₀

where f(n),g(n) >0 , g(n) > f(n)
for n>=0 .
if yes, find constant c, n₀

from (1), we have

cg(n)-(c+2) f(n) >= 0 (a), or we can rewrite as

c( g(n)-f(n) ) -2f(n) >=0 (b)

a) c= 0 , -2 false
c<0 , false
c > 0 , ?

b)g-f > 0
c<=0, false
c=-2, false
c>0 ?

don't know how to get around this. any clues?

 

[micromass]

You'll need to find a c such that

c\geq\frac{2f(n)}{g(n)-f(n)}

Maybe you can take the limit of the right-hand side and show that it is not infinite...

 

[atrus_ovis]

How will that information help?

 

[micromass]

That's worth poundering about...