Comparison Test and Geometric Series

[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 on a phone haha

 

[HallsofIvy]

Your "attempt" seems to be missing something- like any attempt at all! There are a number of ways to show that "a< b". One is to show that b/a> 1. another is to show that b-a> 0.