Comparison Test and Geometric Series

The Problem:

Let a(n) = (n^2)/(2^n)

Prove that if n>=3, then:

(a(n+1))/(a(n)) <= 8/9

By using this inequality for n = 3,4,5,..., prove that:

a(n+3) <= ((8/9)^n)(a(3))

Using the comparison test and results concerning the convergence of the geometric series, show that:

The SUM(from n=1 to infinity) of a(n+3) is convergent.

Now use the shift rule to show that:

The SUM(from n=1 to infinity) of a(n) is convergent.

Attempt:

Shown that if you let n=3 then (a(n+1))/(a(n)) = 8/9

How do I then show that for all n>=3, (a(n+1))/(a(n)) <= 8/9

From then on I'm not quite sure :(

thanks for any help and sorry for the lack of LaTex, I'm on a phone haha

Re: Comparison Test and Geometric Series

Quote:

Originally Posted by

**lmstaples** The Problem:

Let a(n) = (n^2)/(2^n)

Prove that if n>=3, then:

(a(n+1))/(a(n)) <= 8/9

By using this inequality for n = 3,4,5,..., prove that:

a(n+3) <= ((8/9)^n)(a(3))

Using the comparison test and results concerning the convergence of the geometric series, show that:

The SUM(from n=1 to infinity) of a(n+3) is convergent.

Now use the shift rule to show that:

The SUM(from n=1 to infinity) of a(n) is convergent.

Attempt:

Shown that if you let n=3 then (a(n+1))/(a(n)) = 8/9

How do I then show that for all n>=3, (a(n+1))/(a(n)) <= 8/9

From then on I'm not quite sure :(

thanks for any help and sorry for the lack of LaTex, I'm on a phone haha

Notice that . So

and

For your statement that for all , you need to be able to show that for all . So let's test this inequality and see if it is true for all ...

Obviously we can disregard the negative answer, and notice that , so .

So your inequality is true for all , and since you have shown that when , it follows that for all .