Determine if the SERIES converges or DIVERGES

[shamieh]

Determine if the positive term series is convergent or divergent
\(\displaystyle
\sum^{\infty}_{n = 0} \frac{(2n + 3)^2}{(n + 1)^3}\)

So what I did was look at it as 2n + 3/n + 1 , then I did the limit as n --> infinity using lopitals to get 2/1 which = 2. Then i said this must be divergent since lim a_n as n -> infty is != to 0 by the nth term test this diverges. Would that be correct?

 

[Prove It]

Determine if the positive term series is convergent or divergent
\(\displaystyle
\sum^{\infty}_{n = 0} \frac{(2n + 3)^2}{(n + 1)^3}\)

So what I did was look at it as 2n + 3/n + 1 , then I did the limit as n --> infinity using lopitals to get 2/1 which = 2. Then i said this must be divergent since lim a_n as n -> infty is != to 0 by the nth term test this diverges. Would that be correct?

Checking the limit of the terms is always a good start, because like you said, if this limit is not 0 then the series is divergent. But you are taking the wrong limit. The terms are $\displaystyle \begin{align*} \frac{(2n+3)^2}{(n+1)^3} \end{align*}$, not $\displaystyle \begin{align*} \frac{2n+3}{n+1} \end{align*}$. Using the correct terms we have:

$\displaystyle \begin{align*} \frac{(2n+3)^2}{(n+1)^3} &= \frac{4n^2 + 12n + 9}{n^3 + 3n^2 + 3n + 1} \\ &= \frac{\frac{4}{n} + \frac{12}{n^2} + \frac{9}{n^3}}{1 + \frac{3}{n} + \frac{3}{n^2} + \frac{1}{n^3}} \\ &\to \frac{0 + 0 + 0}{1 + 0 + 0 + 0} \\ &= \frac{0}{1} \\ &= 0 \end{align*}$

So the terms actually do go to 0. That means you will need to apply another test to try to determine the convergence of this series. I would suggest the RATIO test.

 

[shamieh]

using ratio I obtained: \(\displaystyle \frac{(2n + 4)^2}{(n + 2)^3} * \frac{(n + 1)^3}{(2n + 3)^2}\) as n---> \(\displaystyle \infty\) = 200/432 so by ratio test L < 1 so the Series converges correct?

 

[Prove It]

using ratio I obtained: \(\displaystyle \frac{(2n + 4)^2}{(n + 2)^3} * \frac{(n + 1)^3}{(2n + 3)^2}\) as n---> \(\displaystyle \infty\) = 200/432 so by ratio test L < 1 so the Series converges correct?

Your ratio is incorrect. $\displaystyle \begin{align*} a_n = \frac{(2n+3)^2}{(n+1)^3} \end{align*}$, so $\displaystyle \begin{align*} a_{n + 1} = \frac{[2(n+1) + 3]^2}{(n+1+1)^3} = \frac{(2n + 5)^2}{(n+2)^3} \end{align*}$, so the ratio is actually

$\displaystyle \begin{align*} \frac{a_{n + 1}}{a_n} = \frac{(2n + 5)^2}{(n + 2)^3} \cdot \frac{ (n + 1) ^3}{(2n + 3)^2} \end{align*}$

Now see what happens to this ratio as $\displaystyle \begin{align*} n \to \infty \end{align*}$.

 

[shamieh]

Ok , fixing my error... I got 288/432... But that can't be right...What am I doing wrong here..

I'm saaying that \(\displaystyle \frac{(2n + 5)^2}{(2n + 3)^2} = 36/16\)

also I'm saying that \(\displaystyle \frac{(n+1)^3}{(n+2)^3} = 8/27\)

then getting 288/432...Am I doing algebra incorrectly? I think that's the casE but i don't see my mistake...

 

[Prove It]

Ok , fixing my error... I got 288/432... But that can't be right...What am I doing wrong here..

I'm saaying that \(\displaystyle \frac{(2n + 5)^2}{(2n + 3)^2} = 36/16\)

also I'm saying that \(\displaystyle \frac{(n+1)^3}{(n+2)^3} = 8/27\)

then getting 288/432...Am I doing algebra incorrectly? I think that's the casE but i don't see my mistake...

Not even close I'm afraid.

$\displaystyle \begin{align*} \frac{(2n + 5)^2}{(2n + 3)^2} &= \left( \frac{2n + 5}{2n + 3} \right) ^2 \\ &= \left( \frac{2n + 3}{2n + 3} + \frac{2}{2n + 3} \right) ^2 \\ &= \left( 1 + \frac{2}{2n + 3} \right) ^2 \\ &\to \left( 1 + 0 \right) ^2 \textrm{ as } n \to \infty \\ &= 1 \end{align*}$

What about the other one?

 

[shamieh]

wow...\(\displaystyle \frac{(n+1)^3}{(n+2)^3} = n/n+2 = ((1+0) + (0))^3 = 1^3\)?? I have no idea..

doesn't make sense because then you would have 1/1 = 1...and test is inconclusive...

 

[Prove It]

wow...\(\displaystyle \frac{(n+1)^3}{(n+2)^3} = n/n+2 = ((1+0) + (0))^3 = 1^3\)?? I have no idea..

doesn't make sense because then you would have 1/1 = 1...and test is inconclusive...

PLEASE be careful with your algebra. $\displaystyle \begin{align*} \frac{ (n + 1)^3}{( n +2)^3} \end{align*}$ is NOT $\displaystyle \begin{align*} \frac{n}{n + 2} \end{align*}$. Try again.

But you are correct that you should end up with the ratio test being inconclusive (I didn't realize this when I suggested it).

Wolfram says a comparison is needed. I'm trying to think of a function to compare it to...

 

[shamieh]

using limit comparison can i compare it to 1/n^2

 

[Prove It]

using limit comparison can i compare it to 1/n^2

Close, but it will be of order 1/n, not 1/n^2 (the top has order 2 and the bottom has order 3)...

$\displaystyle \begin{align*} \frac{(2n + 3)^2}{(n + 1)^3} &= \frac{4n^2 + 12n + 9}{ n^3 + 3n^2 + 3n + 1 } \\ &= \frac{4 + \frac{12}{n} + \frac{9}{n^2}}{n + 3 + \frac{3}{n} + \frac{1}{n^2}} \end{align*}$

For large values of n, this behaves like $\displaystyle \begin{align*} \frac{4}{n + 3} \end{align*}$, which is a harmonic series, well known to be DIVERGENT.

So your series is divergent too :)