Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit question related to Euler's constant

  1. Oct 10, 2013 #1
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 10, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper

    For one, n does not "reach" infinity; nor does 1/n reach zero - it just approaches it (as closely as you want it, by choosing n large enough).

    Secondly, if you take a limit of n, which is formally what you need to do, you cannot have the limit depend on n. It makes as little sense as saying $$\lim_{n \to 2} n x^n = n x^2.$$

    Thirdly, note that 1 + 1/n is strictly larger than 1 for any value of n. If you would take the limit of ##a^n## as ##n \to \infty##, then if a > 1 you would get infinity, and then you can let a go to 1 all you want, you will have a (infinitely) different result.
    So you really need to look at the two occurrences of n "at the same time" - you can't take a part of the limit first and then take another part of the limit. To give an easier example, consider $$\lim_{n \to \infty} n \cdot \frac{1}{n}.$$ Would you agree if I claimed that 1/n goes to zero, so you take the limit of ##n \cdot 0 = 0## which is clearly zero?
  4. Oct 10, 2013 #3
    I am not sure what you are saying here. Can you please explain a bit more?

    Also, how does one go from the limit to the Euler's constant?
    Is there are website where I can see the calculation?

  5. Oct 10, 2013 #4
    Limits are used to get as close to a "breaking point" as possible to observe what happens there without actually hitting it.

    so you have your definition of e:
    $$e = \lim_{n \to \infty} (1 + \frac{1}{n})^n$$

    Since this definition would break down at n is infinity you approach it from as close to that as you can get and therefore use a limit.

    As for a method of calculating it, you use the Taylor series as expressed at the top of that Wikipedia article:
    $$e = \sum_{n = 0}^{\infty} \frac{1}{n!}$$

    This is found using the original definition of e and using the following manipulations:
    By the Binomial Theorem we get
    $$\lim_{n \to \infty} (1 + \frac{1}{n})^n = \lim_{n \to \infty} \sum_{m = 0}^{n} (_m^n) \frac{1}{n^m}$$
    Recalling that $$(_m^n) = \frac{n!}{m!(n-m)!}$$ This is
    $$\lim_{n \to \infty} (1 + \frac{1}{n})^n = \lim_{n \to \infty} \sum_{m = 0}^{n} \frac{n!}{m!(n-m)!} \frac{1}{n^m} = \lim_{n \to \infty} \sum_{m = 0}^{n} \frac{n!}{n^m(n-m)!} \frac{1}{m!}$$
    Now we simply have to show that
    $$ \lim_{n \to \infty} \frac{n!}{n^m(n-m)!} = 1$$ for any m. We do this by cancelling factorials and writing this as
    $$ \lim_{n \to \infty} \frac{\prod_{k = 0}^{m - 1} n-k}{n^m}$$
    Now for the magic, using infinity for n. Because the top and bottom expressions have m terms they will both be mth degree polynomials and when dividing 2 of these, the limit at infinity is the quotient of the leading terms because all other terms become insignificant at such scales. Therefore, this equals one and infinity can be substituted for n in the summation because it rapidly converges and 1/n! will effectively be valued at 0. So, again,
    $$e = \sum_{n = 0}^{\infty} \frac{1}{n!}$$
  6. Oct 11, 2013 #5
    This part is not clear to me. I don't understand how you went from


    = \lim_{n \to \infty} \sum_{m = 0}^{n}\frac{1}{m!}


    [itex]e = \sum_{n = 0}^{\infty} \frac{1}{n!}[/itex]
  7. Oct 11, 2013 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    He switched the name "m" to "n" in the dummy variable, you could as easily say that
  8. Oct 11, 2013 #7
    Thanks. Also, do you know what Compuchip was trying to say in the quote in post #3?
  9. Oct 11, 2013 #8


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    The point is that the formal expression [itex]1^{\infty}[/itex] is an equally indeterminae expression as the more well-known [itex]\frac{0}{0}[/itex]
    How to evaluate such expressions, requiring due care.

    Take for example the expression [itex]\frac{2x}{x}[/itex], which obviously equals 2 for every non-zero x we insert in the expression! But, if we put in x=0, we get the result [itex]\frac{0}{0}[/itex]

    But, if you look at ANOTHER expression, say, [itex]\frac{x}{x}[/itex] this equals 1, rather than 2, for every choice of non-zero x. But, if we insert x=0 here, we ALSO get that expression [itex]\frac{0}{0}[/itex]

    Does 0/0=0/0?

    It doesn't seem so! Similar problems relates to [itex]1^{\infty}[/itex] as well..
  10. Oct 11, 2013 #9
    Yes, he was saying the result of the limit could not depend on a variable defined by the limit.

    He used this as an example
    $$\lim_{n \to 2} n x^n = n x^2$$
    And was saying that it didn't work because the n in nx2 is defined in the limit.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook