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Wow, what am i doing wrong here?

  • #1
217
0
Determine if the series converges absolutely, conditionally, or if it diverges. Give justifications for your answers.

(note: "E" is the sigma notation)

1. E from 1 to infinity of ln(n)/n3

Ok, i basically used the limit comparison test and pulled out a 1/n3. As we know by p-series, 1/n3 converges and now becomes our "bn."

So, using the LCT i get... limit going to infinity of an/bn = lim to infinity of [ln(n)/n3]/(1/n3) which results in the lim to infinity of ln(n) which = infinity. Because our limit is infinity it means that if bn diverges, an diverges as well. But in this case, bn converged! Any thoughts??

Because bn converged i thought the answer would be "converges conditionally" but i was marked points off because the answer is "converges absolutely."
 

Answers and Replies

  • #2
307
3
Determine if the series converges absolutely, conditionally, or if it diverges. Give justifications for your answers.

(note: "E" is the sigma notation)

1. E from 1 to infinity of ln(n)/n3

Ok, i basically used the limit comparison test and pulled out a 1/n3. As we know by p-series, 1/n3 converges and now becomes our "bn."

So, using the LCT i get... limit going to infinity of an/bn = lim to infinity of [ln(n)/n3]/(1/n3) which results in the lim to infinity of ln(n) which = infinity. Because our limit is infinity it means that if bn diverges, an diverges as well. But in this case, bn converged! Any thoughts??

Because bn converged i thought the answer would be "converges conditionally" but i was marked points off because the answer is "converges absolutely."
I would just show that,

[tex]\log n < n \implies \frac{\log n}{n^3} < \frac{1}{n^2}.[/tex]

What do you know about the latter series?
 
  • #3
217
0
Coto, how does that prove the series converges absolutely? And why did you pull a 1/n^2 out instead of a 1/n^3
 
  • #4
307
3
Hey IntegrateMe,

Well first off, what does it mean for a series to absolutely converge?

Second, what do you know about:

[tex]\sum_1 ^{\infty} \frac{1}{n^2}.[/tex]

Does it converge?

Lastly,
[tex]\log n < n \implies \frac{\log n}{n^3} < \frac{n}{n^3} = \frac{1}{n^2},[/tex]

and for [tex]n \geq 1, \log n \geq 0.[/tex]

Hopefully this leads you in the right direction.
 
  • #5
217
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Thanks Coto!

I was just trying to gain some intuition; you helped a lot and i understand convergence a lot better now.
 

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