# Thread: Comparison Test and Geometric Series

1. ## 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

2. ## Re: Comparison Test and Geometric Series

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 \displaystyle \begin{align*} a_n = \frac{n^2}{2^n}, a_{n+1} = \frac{(n+1)^2}{2^{n+1}}, a_{n+2} = \frac{(n+2)^2}{2^{n+2}} \end{align*}. So

\displaystyle \begin{align*} \frac{a_{n+1}}{a_n} &= \frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}} \\ &= \frac{2^n \left( n + 1 \right)^2}{2^{n+1}\, n^2} \\ &= \frac{n^2 + 2n + 1}{2n^2} \end{align*}

and

\displaystyle \begin{align*} \frac{a_{n+2}}{a_n} &= \frac{\frac{(n+2)^2}{2^{n+2}}}{\frac{(n+1)^2}{2^{n +1}}} \\ &= \frac{2^{n+1}\left( n + 2 \right)^2}{2^{n+2}\left( n + 1 \right)^2} \\ &= \frac{n^2 + 4n + 4}{2n^2 + 4n + 2} \end{align*}

For your statement that \displaystyle \begin{align*} \frac{a_{n+1}}{a_n} \leq \frac{8}{9} \end{align*} for all \displaystyle \begin{align*} n \geq 3 \end{align*}, you need to be able to show that \displaystyle \begin{align*} \frac{a_{n+2}}{a_{n+1}} \leq \frac{a_{n+1}}{a_n} \end{align*} for all \displaystyle \begin{align*} n \geq 3 \end{align*}. So let's test this inequality and see if it is true for all \displaystyle \begin{align*} n \geq 3 \end{align*}...

\displaystyle \begin{align*} \frac{a_{n+2}}{a_{n+1}} &\leq \frac{a_{n+1}}{a_n} \\ \frac{n^2 + 4n + 4}{2n^2 + 4n + 2} &\leq \frac{n^2 + 2n + 1}{2n^2} \\ 2n^2 \left( n^2 + 4n + 4 \right) &\leq \left(n^2 + 2n + 1\right) \left( 2n^2 + 4n + 2 \right) \\ 2n^4 + 8n^3 + 8n^2 &\leq 2n^4 + 8n^3 + 12n^2 + 8n + 2 \\ 0 &\leq 4n^2 + 8n + 2 \\ 0 &\leq n^2 + 2n + \frac{1}{2} \\ 0 &\leq n^2 + 2n + 1^2 - 1^2 + \frac{1}{2} \\ 0 &\leq \left( n + 1 \right)^2 - \frac{1}{2} \\ \frac{1}{2} &\leq \left( n + 2 \right)^2 \\ \left| n + 2 \right| &\geq \frac{\sqrt{2}}{2} \\ n + 2 \leq -\frac{\sqrt{2}}{2} \textrm{ or } n + 2 &\geq \frac{\sqrt{2}}{2} \\ n \leq \frac{-2 - \sqrt{2}}{2} \textrm{ or } n &\geq \frac{-2 + \sqrt{2}}{2} \end{align*}

Obviously we can disregard the negative answer, and notice that \displaystyle \begin{align*} \sqrt{2} < \sqrt{4} \end{align*}, so \displaystyle \begin{align*} \frac{-2 + \sqrt{2}}{2} < \frac{-2 + 2}{2} = 0 \end{align*}.

So your inequality \displaystyle \begin{align*} \frac{a_{n+2}}{a_{n+1}} \leq \frac{a_{n+1}}{a_n} \end{align*} is true for all \displaystyle \begin{align*} n \geq 0 \end{align*}, and since you have shown that \displaystyle \begin{align*} \frac{a_{n+1}}{a_n} = \frac{8}{9} \end{align*} when \displaystyle \begin{align*} n = 3 \end{align*}, it follows that \displaystyle \begin{align*} \frac{a_{n+1}}{a_n} \leq \frac{8}{9} \end{align*} for all \displaystyle \begin{align*} n \geq 3 \end{align*}.