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Limit question

  1. Oct 3, 2007 #1
    lim(n to inf)[e^n-(1+1/n)^n^2]
    after some calculations [e^n-e^(n-1/2+1/(3n)+o(1/n))] *
    then i took e^n out,so e^n(1-e^(-1/2+1/(3n)+o(1/n))) and i expanded second term
    e^n(1-1+1/2-1/(3n)+o(1/n)),so the limit is inf.But my teacher said that my calculations up to * is true then all calculations are wrong.
  2. jcsd
  3. Oct 3, 2007 #2


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    That's hard to read. Why don't you try typing it out in LaTeX?
  4. Oct 4, 2007 #3
    looks like you are trying to find
    [tex]\lim_{n \rightarrow\infty} [e^n-(1+\frac{1}{n})^{n^2}}][/tex]
    If so have you tried using the fact that [tex]n^2=n*n[/tex]
    and looking at what happens to
    [tex]\lim_{n \rightarrow\infty} (1+\frac{1}{n})^{n}[/tex]
  5. Oct 4, 2007 #4
    than actually answer is 0.Yes?
  6. Oct 4, 2007 #5
    [tex]\lim_{n \rightarrow\infty}[(1+\frac{1}{n})^n}]=e^n[/tex]
    [tex]\lim_{n \rightarrow\infty} [e^n-(1+\frac{1}{n})^{n^2}}]=e^n-e^n=0[/tex]

    Is it true?
  7. Oct 4, 2007 #6
    but my teacher said that my solutions up to
    Here first part is tends to infinity faster.Then is it infinity?
  8. Oct 4, 2007 #7
    I couldn't see how yo got that expansion but even a direct expansion of the bracketed term still gives you just exp(n) so I think the answer is just zero.
  9. Oct 5, 2007 #8
    Today my prof.said that the answer is really infinity.
  10. Oct 5, 2007 #9
    I would like to see that proof.
  11. Oct 5, 2007 #10


    this one tends to
    Last edited: Oct 5, 2007
  12. Oct 5, 2007 #11
    Thank you. Very interesting result and still not obvious to me.
  13. Oct 7, 2007 #12

    Gib Z

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    Quite a clever method.

    SanjeevGupta- What he did was simply note that since the exponential and logarithm are inverse functions, [tex]x = e^{\ln x}[/tex]. Then he also used the result [tex]\log (k^n) = n \log k[/tex], and then expanded the log with its taylor series.
  14. Oct 7, 2007 #13
    Thanks Gib Z. I now realise that since [tex]e^n[/tex] is an infinitely long series to start with and whatever n we chose [tex](1+\frac{1}{n})^{n^2}[/tex] has a finite number of terms so there's always going to be an infinitely long string of terms left after the subtraction starting with the [tex]n^2+2[/tex] term of [tex]e^n[/tex].
    So this reminds me to be very careful when looking at limits of expressions that involve infinite series.
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