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Proof that e is irrational using Taylor series

  1. Jan 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Using the equality ##e = \sum_{k=0}^n \frac{1}{k!} + e^\theta \frac{1}{(n+1)!}## with ##0< \theta < 1##, show the inequality ##0 < n!e-a_n<\frac{e}{n+1}## where ##a_n## is a natural number.
    Use this to show that ##e## is irrational.
    (Hint: set ##e=p/q## and ##n=q##)

    2. Relevant equations
    N/A

    3. The attempt at a solution
    I think I manage to do the first part. Multiply both sides with ##n!##
    ##n!e = \sum_{k=0}^n \frac{n!}{k!} + \frac{e^\theta}{n+1}##
    Setting ##\theta = 0## and ##\theta=1## respectively we have the inequality
    ##0 < n!e-a_n<\frac{e}{n+1}## with ##a_n = \frac{n!}{k!}##.

    For the second part assume ##e= \frac{p}{q}## where ##p,q## are natural numbers without a common divisor (except 1).
    Set ##n=q## in the above inequality
    ##0 < q!\frac{p}{q}-a_q < \frac{p}{q(q+1)}##
    Since ##q,p## and ##q+1## is positive we can multiply (divide) by these
    ##0 < (q+1)(q!-qa_q/p)< 1##
    My hopes at this point would be to have an integer in the middle and hence get a contradiction (since there is no integers between 0 and 1), but that ##p## is in the way.
     
  2. jcsd
  3. Jan 18, 2016 #2

    Samy_A

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    From ##0 < q!\frac{p}{q}-a_q < \frac{p}{q(q+1)}##, you get
    ##0<p(q-1)!-a_q<\frac{e}{q+1}##.
    I you could show that ##\frac{e}{q+1}<1##, you are done.
     
  4. Jan 18, 2016 #3
    Thanks! We know that ##2 < e < 3## (this is easy to prove using the limit definition of e). If ##q \ge 2## then the inequality is obviously true since ##e < 3## and hence we have a contradiction.

    The case for ##q=1## means we have a natural number so it's enough to show it for ##q \ge 2## as above.
     
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