Thread: Series convergent by AST, divergent by LCT?

1. Series convergent by AST, divergent by LCT?

Hello,

I have a question regarding the following problem:

Describe if the following series converges absolutely, conditionally or diverges.

Sigma (n=2 to infinity) (-1)^n * (n^2+2n)/(n^3-3n))

Using the "absolute" ratio test, I get lim = 1, so no conclusion can be drawn.

Using the "absolute" limit comparison test with 1/n, I get lim = 1, concluding that it diverges. (as 1/n diverges)

Using the alternating series test this series a) is decreasing and b) the limit goes to 0; concluding that this series converges.

As the absolute part of the series diverges by the limit comparison test, but the absolute part of the series converge by the alternating series test, can I conclude this series converges conditionally?

Thanks

2. Originally Posted by FlyingDan
Hello,

I have a question regarding the following problem:

Describe if the following series converges absolutely, conditionally or diverges.

Sigma (n=2 to infinity) (-1)^n * (n^2+2n)/(n^3-3n))

Using the "absolute" ratio test, I get lim = 1, so no conclusion can be drawn.

Using the "absolute" limit comparison test with 1/n, I get lim = 1, concluding that it diverges. (as 1/n diverges)

Using the alternating series test this series a) is decreasing and b) the limit goes to 0; concluding that this series converges.

As the absolute part of the series diverges by the limit comparison test, but the absolute part of the series converge by the alternating series test, can I conclude this series converges conditionally?

Thanks
Absolute convergence: Use the comparison test. Note that $\frac{n^2+2}{n^3 - 3} > \frac{n^2}{n^3} = \frac{1}{n}$.

Originally Posted by mr fantastic
Absolute convergence: Use the comparison test. Note that $\frac{n+2}{n^2 - 3} > \frac{n}{n^2} = \frac{1}{n}$.
Am I missing something here? The series I was referring to is $\frac{n^2+2}{n^3 - 3}$.

4. Originally Posted by FlyingDan

Am I missing something here? The series I was referring to is $\frac{n^2+2}{n^3 - 3}$.
I misread the expression. I've edited my reply. The argument and result remain exactly the same.

5. Originally Posted by mr fantastic
I misread the expression. I've edited my reply. The argument and result remain exactly the same.
I'm still a bit lost though. Doesnt 1/n diverge as it is a p series with p=1?

6. Originally Posted by FlyingDan
Hello,

I have a question regarding the following problem:

Describe if the following series converges absolutely, conditionally or diverges.

Sigma (n=2 to infinity) (-1)^n * (n^2+2n)/(n^3-3n))

Using the "absolute" ratio test, I get lim = 1, so no conclusion can be drawn.

Using the "absolute" limit comparison test with 1/n, I get lim = 1, concluding that it diverges. (as 1/n diverges)

Using the alternating series test this series a) is decreasing and b) the limit goes to 0; concluding that this series converges.

As the absolute part of the series diverges by the limit comparison test, but the absolute part of the series converge by the alternating series test, can I conclude this series converges conditionally?

Thanks
As Mr. Fantastic noted $\sum_{n=1}^{\infty}\left|\frac{(-1)^n(n^2+2n)}{n^3-3}\right|=\sum_{n=1}^{\infty}\frac{n^2+2n}{n^3-3}$ and that last series diverges by limit comparison to the harmonci series. But note that if $a_n=\frac{n^2+n}{n^3-3}$ that $a_{n+1}\leqslant{a_n}$ and $\lim_{n\to\infty}a_n=0$, so $\sum_{n=1}^{\infty}\frac{(-1)^n(n^2+2n)}{n^3-3}$ converges by the Leibniz's Criterion (AST). So we conclude that $\sum_{n=1}^{\infty}\frac{(-1)^n(n^2+2n)}{n^3-3}$ converges conditionally.

7. Originally Posted by FlyingDan
I'm still a bit lost though. Doesnt 1/n diverge as it is a p series with p=1?
Yes it does . And if you're familiar with the comparison test you will know that if $u_n > v_n$ and $\sum v_n$ diverges then $\sum u_n$ diverges.