Limit question related to Euler's constant

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  • #2
CompuChip
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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?
 
  • #3
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Thanks.
Secondly, if you take a limit of n, which is formally what you need to do, you cannot have the limit depend on n.
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?

Thanks.
 
  • #4
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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,
$$e = \sum_{n = 0}^{\infty} \frac{1}{n!}$$
 
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  • #5
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Thanks.
infinity can be substituted for n in the summation because it rapidly converges and 1/n! will effectively be valued at 0.
This part is not clear to me. I don't understand how you went from

[itex]

= \lim_{n \to \infty} \sum_{m = 0}^{n}\frac{1}{m!}
[/itex]

to

[itex]e = \sum_{n = 0}^{\infty} \frac{1}{n!}[/itex]
 
  • #6
arildno
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He switched the name "m" to "n" in the dummy variable, you could as easily say that
[tex]e=\sum_{m=0}^{\infty}\frac{1}{m!}[/tex]
 
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  • #7
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Thanks. Also, do you know what Compuchip was trying to say in the quote in post #3?
 
  • #8
arildno
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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..
 
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  • #9
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Thanks. Also, do you know what Compuchip was trying to say in the quote in post #3?
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.
 

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