# Series: What do I have the right to do?

1. Jan 15, 2009

### duffman

1. The problem statement, all variables and given/known data
I have to say if the series converge or diverge.

$$\infty$$
$$\sum$$ LN(n$$/2n+5$$)
n=1

2. Relevant equations

I found this series to be divergent. However, I dont know if the way I proceeded is correct.

3. The attempt at a solution

I first separated the ln into two lns ( ln(n) - ln(2n+5 ). Then, I used it as a power of e to remove the lns, thus being left with -n -5. I then went for the limit divergence test, and it ended up diverging.

Do I have the right to proceed that way?

2. Jan 15, 2009

### Dick

Are you summing ln(n/(2n+5))? If so, the first think you should check is whether the sequence ln(n/(2n+5)) approaches 0 as n->infinity. Does it?

3. Jan 15, 2009

### duffman

Well ...

ln(n) - ln(2n+5) equals infinity minus infinity, which is an undetermined form. from there I'm just lost. I remembered about the e trick, but I'm not certain if I can use it in the case of series. This is probably very basic stuff but I havent done it in ages... I totally lost my "instinct". I dont want to waste anyone's time...

4. Jan 15, 2009

### Dick

Whats the limit of n/(2n+5) as n->infinity? Don't split it first.

5. Jan 15, 2009

### duffman

That would be infinity on infinity... If I use L'Hopital's rule, that would give me 1/2.

6. Jan 15, 2009

### NoMoreExams

You don't need to L'Hopitals. That's killing a cockroach with a nuke. Divide top and bottom by n.

7. Jan 15, 2009

### duffman

Hahaha. Alright. Resulting in 1/(2+5/n) and 5/n goes to zero.

That gives us the limit of a constant, which would then be ln(1/2) ?

8. Jan 15, 2009

### AEM

Correct. And if you add up an infinite number of those ln(1/2)'s, what do you get?

9. Jan 15, 2009

### duffman

So since the limit does not equal zero, the series is divergent.

10. Jan 15, 2009

### Dick

That is so correct.

11. Jan 16, 2009

### duffman

Thanks everyone!