Does the series \(\sum_{n=2}^\infty\frac{(-1)^n}{\ln(n)}\) converge?

[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.

 

[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.

 

[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)

 

[STEMucator]

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 $|a_{n+1}| ≤ |a_n|$

Then your series converges.

 

[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.

 

[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?

 

[SammyS]

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.

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 ?

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

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

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?

Yes, that's correct for that series, but as stated before, you can't use the comparison test for alternating series.