1. ## converge or diverge.

Does this equation converge absolutely, converge conditionally, or does it diverge?

heres the equation: http://img219.imageshack.us/img219/8247/untitled2qf.jpg

Thanks for any help with this one.

2. Originally Posted by rcmango
Does this equation converge absolutely, converge conditionally, or does it diverge?

heres the equation: http://img219.imageshack.us/img219/8247/untitled2qf.jpg

Thanks for any help with this one.
Use the alternating series test.

To determine if $\displaystyle \sum_{n=2}^{\infty} (-1)^n \frac{n}{1+n \ln n}$ converges, set $\displaystyle a_n = \frac{n}{1+n \ln n}$.

You can see that $\displaystyle a_n$ decreases as $\displaystyle n$ increases. I'll demonstrate by showing if it were a function on the reals, its derivative would be negative on the relevant domain:

$\displaystyle \frac{d}{dn} a_n = \frac{1 - n}{(1 + n \ln n)^2}$, which is negative for $\displaystyle n \ge 2$, which is where we're counting from.

The alternating series test also requires that $\displaystyle \lim_{n\to\infty} a_n = 0$. This is true and easy to verify using L'Hospital's rule.

The conditions of the alternating series test are satisfied. The series converges.

3. I forgot to answer part of the question.
Does it converge conditionally or absolutely?

Well consider the harmonic series, which is $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$. This is known to diverge.

Suppose for the sake of argument that your series converged absolutely (there may be a simpler way of proving this, but this came to mind first. You could try integrating, but it looks challenging).

Then at least $\displaystyle a_n \le \frac{1}{n}$ for all $\displaystyle n \ge 2$.

$\displaystyle 1 + n \ln n \ge n^2$.
$\displaystyle n \ln n \ge n^2 - 1$.

Since $\displaystyle n \ge 1$, the following arrangement is valid:

$\displaystyle n \ge e^{(n - \frac{1}{n})}$.

It should be clear that for large $\displaystyle n$, this relationship fails.

Thus, in fact $\displaystyle a_n > \frac{1}{n}$ for all $\displaystyle n \ge 2$.
So it cannot converge absolutely, since each term is larger than the terms of a series known to diverge.

Since the alternating series converges as proved in my first response, then we now know it only converges conditionally.

4. thankyou for the help so far, which part of the equation do i prove its the alternating series with l'hopitals rule?

thankyou.

5. Originally Posted by rcmango
which part of the equation do i prove its the alternating series with l'hopitals rule?
.
YOu show that the limit of the sequence is zero.

6. Originally Posted by rcmango
thankyou for the help so far, which part of the equation do i prove its the alternating series with l'hopitals rule?

thankyou.
Well, you need to show that $\displaystyle \lim_{n\to\infty} \frac{n}{1+n \ln n} = 0$.

If you plug $\displaystyle \infty$ directly into the expression, you'll get $\displaystyle \frac{\infty}{\infty}$ which is an undetermined form, so you can use L'Hospital's rule.

Differentiate the numerator and denominator and evaluate the much simpler:
$\displaystyle \lim_{n\to\infty} \frac{1}{\ln n + 1}$.

If you know that logarithms increase to infinity, it should be clear that this limit is 0, which is what we wanted to show for that step of the alternating series test.

7. Okay I see it now, Thankyou for all the help.