Series: What do I have the right to do?

  • Thread starter duffman
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In summary, the conversation discusses whether a series converges or diverges, specifically the series ln(n/(2n+5)). After attempting to use the limit divergence test, it is determined that the series is divergent due to the limit of ln(n/(2n+5)) not approaching zero. This is confirmed by adding up an infinite number of ln(1/2)'s, resulting in a divergent series.
  • #1
duffman
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Homework Statement


I have to say if the series converge or diverge.

[tex]\infty[/tex]
[tex]\sum[/tex] LN(n[tex]/2n+5[/tex])
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?
 
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  • #2
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
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...
 
  • #4
Whats the limit of n/(2n+5) as n->infinity? Don't split it first.
 
  • #5
That would be infinity on infinity... If I use L'Hopital's rule, that would give me 1/2.
 
  • #6
You don't need to L'Hopitals. That's killing a cockroach with a nuke. Divide top and bottom by n.
 
  • #7
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
Correct. And if you add up an infinite number of those ln(1/2)'s, what do you get?
 
  • #9
So since the limit does not equal zero, the series is divergent.
 
  • #10
duffman said:
So since the limit does not equal zero, the series is divergent.

That is so correct.
 
  • #11
Thanks everyone!
 

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