Series: What do I have the right to do?

[duffman]

Homework Statement

I have to say if the series converge or diverge.

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

Homework Equations

I found this series to be divergent. However, I don't know if the way I proceeded is correct.

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?

 

[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?

 

[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 haven't done it in ages... I totally lost my "instinct". I don't want to waste anyone's time...

 

[Dick]

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

 

[duffman]

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

 

[NoMoreExams]

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

 

[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) ?

 

[AEM]

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

 

[duffman]

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

 

[Dick]

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

That is so correct.

 

[duffman]

Thanks everyone!