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Series- convergent or divergent?

  1. Feb 2, 2008 #1
    from n=1 to infinity

    does the series converge or diverge?

    n!/n^n




    its in the secition of the book with the comparison test and limit comparison test.

    if you compare it with 1/n^n (this is a geomoetric series) you get a= 1/n amd r= 1/n

    but in the thrm r = to some finite number and i dont know if the geometric series thrm applies here.

    help please.:yuck:
     
  2. jcsd
  3. Feb 2, 2008 #2

    StatusX

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    A geometric series has a constant factor between successive elements, so 1/n^n does not qualify. Look at:

    [tex] \frac{n!/n^n}{(n-1)!/(n-1)^{n-1}} [/tex]
     
  4. Feb 2, 2008 #3
    statusx....help me solve my problem rather than leading me to different problems

    advie to a homework helper....

    if geometric series does'nt apply here, what shouudl be done when comparing
     
  5. Feb 2, 2008 #4

    quasar987

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    Re-read StatusX's post. The last part of his message is a major hint.
     
  6. Feb 2, 2008 #5
    hmmm what does (n-1)! = ?

    n!(n-1)?
     
  7. Feb 2, 2008 #6
    It means take n, subtract 1, then apply the factorial operator to the result.
     
  8. Feb 2, 2008 #7

    HallsofIvy

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    No, (n+1)!= n!(n+1) which is what you may be thinking of. It should be obvious that (n-1)! is less than n!.
     
  9. Feb 2, 2008 #8
    I was able to solve this using the limit test and some strategic use of logarithms. Basically, set the limit of the limit test equal to [tex]y[/tex],

    [tex]
    y &= \lim_{n \rightarrow \infty} \left | \left ( \frac{n}{n+1} \right )^n \right |\\
    [/tex]

    take the ln of both sides,

    [tex]
    \ln y &= \lim_{n \rightarrow \infty} n {\ln \left |\frac{n}{n+1}\right | }
    [/tex]

    then use L'Hopitals Rule, and solve for [tex]y[/tex].
     
  10. Feb 2, 2008 #9
    can anyone write the solution?

    please dont whore my thread. i just need the solution to the problem, i'm not lookign for hints.

    thanks
     
  11. Feb 2, 2008 #10
    Unfortunately, unless you show us you're work (i.e. that you've tried what we've suggested), we can't write the whole solution. However, I can tell that yes, it does converge. Oh, and my suggestion used the ratio test, not the limit test. >_<
     
  12. Feb 2, 2008 #11
    oh okay...ratio test. now i see
     
  13. Feb 2, 2008 #12

    Gib Z

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    It seems it wasn't enough to have your own "give me the answer quickly" attitude, but you want to spread it, like in this thread: https://www.physicsforums.com/showthread.php?t=212583

    Why do you think you needed to give advice to a homework helper? They got the badge because they have a somewhat decent knowledge of the problems they give advice on, and know how to help. It they lead you to a different problem, then theres a good chance it will also solve the original one.

    Help is not "whoring a thread". If you just want a solution, do it yourself and quit wasting our time.

    PS. Is there any reason you chose your username to be NuclearRape66?
     
  14. Feb 2, 2008 #13
    are you gonna go through my inbox?

    and what you're doing right there is whoring my thread...if you dont wanan help dont help. dotn speak for everyone, speak for yourself.
     
  15. Feb 2, 2008 #14

    Gib Z

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    We want to help, not give you the bloody answer. We're not a freaking oncall service at here for your petty mathematical needs. I speak for the vast majoirty of the forum users here. Only a few people, like you, find it nessescary to ruin and disobey forum rules.
     
  16. Feb 3, 2008 #15

    HallsofIvy

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    So far you have:

    1. Shown that you cannot do even basic arithmetic.

    2. Insulted people who tried to help you.

    3. Refused to even try yourself.

    I think we would all be better off if you just went somewhere else.
     
  17. Feb 3, 2008 #16
    stop whoring my thread, if you dont like it, then dont bother to click on my thread and waste your time...
     
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