# Homework Help: Limit convegion

1. Dec 15, 2012

### maxitis

Sequence Convergence

$$\sum_{n=2}^\infty\frac{(-1)^n}{ln(n)}$$

I have tried some comparisons bot not conclude:
$$\sum_{n=2}^\infty\frac{(-1)^n}{ln(n)}<=\sum_{n=2}^\infty\frac{(-1)^n}{1}$$
$$\sum_{n=2}^\infty\frac{(-1)^n}{x-1}<=\sum_{n=2}^\infty\frac{(-1)^n}{ln(n)}$$

Somebody having any insights?
Thank you.

Last edited: Dec 15, 2012
2. Dec 15, 2012

### HallsofIvy

Since that is an alternating series, all you have to show is that the sequence of the absolute values of the terms goes to 0.

3. Dec 15, 2012

### maxitis

The absolute value does not converge.
$$\sum_{n=2}^\infty\frac{1}{ln(n)}>=\sum_{n=2}^{\infty}\frac{1}{x-1}$$
So i think we can't use that

(i write right the infinity but it always put a space between)

Last edited: Dec 15, 2012
4. Dec 15, 2012

### Zondrina

Are you familiar with the AST? ( Alternating series test ).

Re-write your summation in the form :

$\sum_{n=2}^{∞} (-1)^n \frac{1}{ln(n)} = \sum_{n=2}^{∞} (-1)^n a_n$

Now if these two conditions are met :

1. If $a_n → 0$ as $n → ∞$
2. and $|a_{n+1}| ≤ |a_n|$

5. Dec 15, 2012

### micromass

Like the others have said, you need to use the alternating series test.

But I would like to add that the comparison tests only work for series with positive terms. If some of the terms are negative, then comparison does not apply.

6. Dec 15, 2012

### maxitis

so the series converges right?
irrelevant, the:
$$\sum_{n=1}^\infty\frac{(-1)^n}{n}=\sum_{n=1}^\infty\frac{1}{2*n}-\frac{1}{2*n-1}$$
isn't that correct?

7. Dec 15, 2012

### SammyS

Staff Emeritus
Re: Series Convergence

What HallsofIvy said was that an alternating series converges if the SEQUENCE of the absolute value of its terms converges to zero.

I your case, does the sequence $\displaystyle \ \ \frac{1}{\ln(2)}\,,\ \frac{1}{\ln(3)}\,,\ \frac{1}{\ln(4)}\,,\ \frac{1}{\ln(5)}\,,\ \frac{1}{\ln(6)}\,,\ \frac{1}{\ln(7)}\dots \ \$ converge to zero ?
It is a very annoying thing which happens to me from time to time !

To fix it, all that I did was put a space before the \infty: { \infty}$$\sum_{n=2}^\infty\frac{1}{ln(n)}>=\sum_{n=2}^{\infty}\frac{1}{x-1}$$$$\sum_{n=2}^\infty\frac{1}{ln(n)}>=\sum_{n=2}^{ \infty}\frac{1}{x-1}$$
Yes, that's correct for that series, but as stated before, you can't use the comparison test for alternating series.

Last edited: Dec 15, 2012