1. ## series convergence

Determine whether the series is convergent or divergent. If it is convergent, find it's sum.

$\displaystyle \sum_{k=1}^\infty\frac{k(k+2)}{(k+3)^3}$

It seems to me that it should converge, because the power of the denominator is greater than the power of the numerator, but I'm having a tough time figuring out how to find the answer.

2. It is divergent. The numerator is quadradic while denominator is cubic, their ratio is simpler behaving like 1/k which is divergent.

3. Use limit comparsion. Compare with $\displaystyle \frac {1}{k}$.

4. As an example, can you post how to show that 1/k diverges?

5. Originally Posted by angel.white
As an example, can you post how to show that 1/k diverges?
What? What level of mathematics are you trying to do?
That is known as the harmonic series.
That is the first divergent series you should have studied.
Go to your textbook and do some studying.

6. Originally Posted by angel.white
As an example, can you post how to show that 1/k diverges?
Always try this first.

7. Originally Posted by ThePerfectHacker
Always try this first.
Thank you, would this be an acceptable answer?

$\displaystyle \sum_{k=1}^\infty\frac{k(k+2)}{(k+3)^3}$

If we evaluate a given term in our sum as it moves towards infinity, we get
$\displaystyle \lim_{n \to \infty}A_n = \lim_{k\to \infty} \frac{k(k+2)}{(k+3)^3}$

$\displaystyle =\lim_{k\to \infty} \frac{k^2+2k}{k^3+9k^2+27k+27}$

$\displaystyle =\lim_{k\to \infty} \frac{1+\frac 2k}{k+9+\frac{27}k+ \frac{27}{k^2}}$

$\displaystyle =\lim_{k\to \infty} \frac{1+\frac 2k}{k+9+\frac{27}k+ \frac{27}{k^2}}$

$\displaystyle =\lim_{k\to \infty} \frac{1+\lim_{k\to \infty} \frac 2k}{\lim_{k\to \infty}k+9+\lim_{k\to \infty}\frac{27}k+ \lim_{k\to \infty}\frac{27}{k^2}}$

(Note that I'm not sure if this step is valid, it seems unlikely that I can evaluate some of the limits while not evaluating others, but if I can, then I think my answer will work)
$\displaystyle =\frac{1}{\lim_{k\to \infty} k}$

$\displaystyle =\lim_{k\to \infty}\frac{1}{k}$

So we can see that as k moves towards infinity, our terms converge to the same values as the terms in the sum

$\displaystyle \sum_{k=1}^\infty \frac 1k$

Which is a divergent series known as the Harmonic series.

Thus our equation is divergent.

8. Originally Posted by angel.white
Thank you, would this be an acceptable answer?

$\displaystyle \sum_{k=1}^\infty\frac{k(k+2)}{(k+3)^3}$

If we evaluate a given term in our sum as it moves towards infinity, we get
$\displaystyle \lim_{n \to \infty}A_n = \lim_{k\to \infty} \frac{k(k+2)}{(k+3)^3}$

$\displaystyle =\lim_{k\to \infty} \frac{k^2+2k}{k^3+9k^2+27k+27}$

$\displaystyle =\lim_{k\to \infty} \frac{1+\frac 2k}{k+9+\frac{27}k+ \frac{27}{k^2}}$

$\displaystyle =\lim_{k\to \infty} \frac{1+\frac 2k}{k+9+\frac{27}k+ \frac{27}{k^2}}$

$\displaystyle =\lim_{k\to \infty} \frac{1+\lim_{k\to \infty} \frac 2k}{\lim_{k\to \infty}k+9+\lim_{k\to \infty}\frac{27}k+ \lim_{k\to \infty}\frac{27}{k^2}}$

(Note that I'm not sure if this step is valid, it seems unlikely that I can evaluate some of the limits while not evaluating others, but if I can, then I think my answer will work)
$\displaystyle =\frac{1}{\lim_{k\to \infty} k}$

$\displaystyle =\lim_{k\to \infty}\frac{1}{k}$

So we can see that as k moves towards infinity, our terms converge to the same values as the terms in the sum

$\displaystyle \sum_{k=1}^\infty \frac 1k$

Which is a divergent series known as the Harmonic series.

Thus our equation is divergent.
Sorry, but all you've really done here is show that $\displaystyle a_k \rightarrow 0$. This is a necessary but not sufficient condition for convergence.

You need to use the comparison test:

$\displaystyle \frac{k(k+2)}{(k+3)^3} > \frac{k(k)}{(k+k)^3} = \frac{k^2}{8k^3} = \frac{1}{8k}$.

So $\displaystyle \sum_{k=1}^\infty\frac{k(k+2)}{(k+3)^3} > \frac{1}{8} \sum_{k=1}^\infty\frac{1}{k}$.

But $\displaystyle \sum_{k=1}^\infty\frac{1}{k}$ diverges. Therefore ......

9. Originally Posted by mr fantastic
Sorry, but all you've really done here is show that $\displaystyle a_k \rightarrow 0$. This is a necessary but not sufficient condition for convergence.

You need to use the comparison test:

$\displaystyle \frac{k(k+2)}{(k+3)^3} > \frac{k(k)}{(k+k)^3} = \frac{k^2}{8k^3} = \frac{1}{8k}$.

So $\displaystyle \sum_{k=1}^\infty\frac{k(k+2)}{(k+3)^3} > \frac{1}{8} \sum_{k=1}^\infty\frac{1}{k}$.

But $\displaystyle \sum_{k=1}^\infty\frac{1}{k}$ diverges. Therefore ......
Thank you, that is a very smart way of doing it. I had tried to do something like this earlier, but I couldn't think of a good function to compare it to, so I tried the method I posted.

10. Originally Posted by angel.white
Thank you, that is a very smart way of doing it. I had tried to do something like this earlier, but I couldn't think of a good function to compare it to, so I tried the method I posted.
The tricks used in increasing the denominator and decreasing the numerator to get a simple comparison are handy ones to remember.

You'd use them in the 'opposite way' if you suspected convergence and therefore were trying to show that a_k < b_k where sum b_k converges.

11. Thanks again, Mr. F! We were looking at ways to perform a comparison test today in class, and I used this method to come up with a good function to compare to, and my instructor seemed to really like that

12. ## THis was already resolved

But I think the best divergence/convergence test is the integral test... if you rewrite $\displaystyle \int_1^∞\frac{k(k+2)}{(k+3)^3}\,dx$ which gives you$\displaystyle \ln(x+3)\$+$\displaystyle frac{8x+21}{2(x+3)^2}\$ evaluated from 0-∞ and as you can see without even evaluating it $\displaystyle \ln(x+3)\$ evaluated at ∞ gives ∞ therefore the integral converges...ergo the summation diverges.

13. Originally Posted by Mathstud28
But I think the best divergence/convergence test is the integral test... if you rewrite $\displaystyle \int_1^∞\frac{k(k+2)}{(k+3)^3}\,dx$ which gives you$\displaystyle \ln(x+3)\$+$\displaystyle frac{8x+21}{2(x+3)^2}\$ evaluated from 0-∞ and as you can see without even evaluating it $\displaystyle \ln(x+3)\$ evaluated at ∞ gives ∞ therefore the integral converges...ergo the summation diverges.
Yes, I like the integral test as well. At the time I asked this, however, we had not covered it yet, it did not appear until the next section (then again, comparison test didn't either, so I'm really not sure what the author was expecting me to do).

14. To be fair to the other tests, the integral test can be a pain at times. It has more strict conditions that must be satisfied in order to use it, and some functions are not particularly easy to integrate.