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Unable to prove e without math. induction

  1. Mar 20, 2009 #1
    1. The problem statement, all variables and given/known data
    How can you prove the following statement without mathematical induction?
    [tex] lim_{n -> \infty} (1 + \frac {1} {n})^{n} = \sum_{n=0}^{\infty} \frac {1} { n! } [/tex]

    3. The attempt at a solution

    The statement can be proven by mathematical induction, but I am interested in how the sum statement can be deduced.
     
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  3. Mar 20, 2009 #2

    Dick

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    I'd be interested in how you prove that with induction. But try expanding (1+1/n)^n in a binomial expansion and look at the limit of the individual terms.
     
  4. Mar 20, 2009 #3
    Let's consider only the first three and the last three terms in the binomial expansion
    [tex] 1 + (n,1) \frac {1} {n} + (n,2) \frac {1} {n^{2}} + (n,3) \frac {1}
    {n^{3}} + ... + (n, n-3) \frac {1} {n^{n-3}} + (n, n-2) \frac {1} {n^{n-2}} +
    (n, n-1) \frac {1} {n^{n-1}} + \frac {1} {n^{n}} [/tex]

    where (n,n) is a combination, for instance.
    (n,0) and (n,n) are one.

    Perhaps, I should now factorise by [tex] \frac {1} {n} [/tex].
     
  5. Mar 20, 2009 #4

    Dick

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    Look at a single term, e.g. (n,3)/n^3. Expand (n,3) in factorials. Cancel some terms in the numerator and denominator. Can you see what happens?
     
  6. Mar 22, 2009 #5
    [tex] 1 + n (\frac {1}{n})^{1}) + n(n - 1) \frac {1}{2!} (\frac {1}{n})^{2}) + n(n - 1)(n - 2) \frac {1}{3!} (\frac {1}{n})^{3}) +
    ... + n(n - 1)(n - 2) \frac {1}{3!} (\frac {1}{n})^{n-3}) + n(n - 1) \frac {1}{2!} (\frac {1}{n})^{n-2}) + n (\frac {1}{n})^{n-1}) + \frac {1} {n^{n}}[/tex]

    Simplifying
    [tex] 2(1 + 1 + \frac {n - 1}{n} \frac {1}{2!} + (n - 1)(n - 2) \frac {1}{n^{2}*3!} +
    ... ) + \frac {1} {n^{n}}[/tex]

    The series diverges, since n is a positive integer.

    We should somehow be able to get the sum notation.
     
    Last edited: Mar 22, 2009
  7. Mar 22, 2009 #6

    Dick

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    You are forgetting the 1/n^k part. For the term containing 3! I get
    (n)(n-1)(n-2)/(n^3*3!). What's the limit of that as n->infinity?
     
  8. Mar 22, 2009 #7
    I get
    [tex]\frac {1 - 3/n - 2/n^2} {3!} => 1/3!,[/tex]
    as n goes to infinity.

    We should now apply the result for the rest of the terms.
     
    Last edited: Mar 22, 2009
  9. Mar 22, 2009 #8

    Dick

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    n/n->1. (n-1)/n->1. (n-2)/n->1. As n->infinity. I get that the limit is 1/3!. Your limit also goes to 1/3!. In spite of a sign error.
     
  10. Mar 22, 2009 #9
    I agree with you.

    I tried to prove the statement unsuccessfully by the proof of contradiction.
    Nevertheless, if we can show that for n = 4 that the sum is 1/4!, we can apparently use mathematical induction to prove the statement.

    I do not see any other way to prove the statement.
     
  11. Mar 22, 2009 #10

    Dick

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    The sum isn't 1/3!. A single term converges to 1/3!. You may be confused because the n on the right side of the equation has nothing to do with the n on the left. Try proving lim n->infinity (1+1/n)^n=sum k=0 to infinity 1/k!.
     
  12. Mar 22, 2009 #11
    Thank you for the clarification!
    I had an idea that there is a relation between n's at both sides.

    The binomial expansion is finally
    [tex] 2 + \frac{n-1}{n*2!} + ... + \frac {1}{n^n}(\frac {n^3(n-1)} {2!} + n^2 + 1) [/tex]

    We know that each of the coefficients such as n/n -> 1, (n-1)/n -> 1 and (n-2)/n -> 1, as n goes to infinity.
    In contrast, the other end of the sequence converges to zero.
    1/n^n -> 0, n^(2-n) -> 0, and n^(3-n)(n-1)/2! -> 0,
    as n goes to infinity.

    We get the following sequence
    [tex] 1 + 1 + 1/2! + 1/3! + ... [/tex]

    Each term in the sequence is in line with the LHS of the equation.
    This completes the proof.

    ---
    Is the proof correct?
     
  13. Mar 22, 2009 #12
    Sorry for entering without permition, for i will not be that helpful either. But, without directly using induction, then it can be shown that:

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


    And also, from the expansion of [tex]f(x)=e^x[/tex] using Taylor series around zero, and also letting x=1, we get


    [tex]e=1+\frac{1}{1!}+\frac{1}{2!}+...=\sum_{n=0}^{\infty}\frac{1}{n!}[/tex]

    so from these two results we have that they are equal.
     
  14. Mar 22, 2009 #13

    Dick

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    I think so. Each term in binomial expansion (n,k)*(1/n)^k converges to 1/k!. So it's plausible that the sum converges to the sum of 1/k! which is e. It's not horribly rigourous, but I've guessing that's what you are supposed to come up with.
     
  15. Mar 22, 2009 #14

    Dick

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    All true. But I think this 'proof' maybe supposed to bypass the knowledge that lim n->infinity (1+1/n)^n=e.
     
  16. Mar 23, 2009 #15
    How would you do the proof more rigorous?
     
  17. Mar 23, 2009 #16

    Dick

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    You'd have to come with with some kind of error estimate for the sum of all of the parts that still contain n. But I wouldn't bother, because as stupidmath said, you know the limit is e and you know that the sum of 1/k! is also a convergent series summing to e from the Taylor series expansion of e^x.
     
  18. Mar 23, 2009 #17

    JJBladester

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    Here is your rigorous proof. Using L'Hôpital's Rule,

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

    Take the natural logarithm of both sides of this equation.

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

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

    [tex]= lim_{n\rightarrow\infty}\left[nln\left(1+\frac{1}{n}\right)\right][/tex]

    [tex]= lim_{n\rightarrow\infty}\left[\frac{ln \left(1+\frac{1}{n}\right)}{\frac{1}{n}}\right][/tex]

    If we try to solve the above limit using direct substitution, we'll obtain the indeterminant form [tex]0/0[/tex]. Thus, we apply L'Hôpital's Rule to attempt to find the limit by taking the derivative of the numerator and denominator and finding the limit of that ratio.

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

    [tex]= lim_{n\rightarrow\infty}\left[\left(\frac{-1/n^{2}}{1+\frac{1}{n}}\right)/\left(-1/n^2\right)}\right][/tex]

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

    Therefore [tex]ln(y) = 1[/tex] which implies y = e by definition of logarithms.

    By comparison, since we let y = the original limit expression, then:

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

    This completes the proof and shows why y converges to e as n gets larger and larger (approaches [tex]\infty[/tex]).
     
  19. Mar 23, 2009 #18
    There is perhaps something wrong in the statements following the expression

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

    The problem seems to be in the derivate.
    I get
    [tex]ln(y) = lim_{n\rightarrow\infty}\left[\left(\frac {1/n} {1 + \frac{1}{n}})\right][/tex]
     
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