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Just wondering...came across this relationship regarding Euler's number in my math tinkerings, but I'm unaware if this particular relationship has a specific name or not:

$$\lim_{x\to\infty}\frac{x^x}{(x-1)^{x-1}} - \frac{(x-1)^{x-1}}{(x-2)^{x-2}} = e$$

So does it have a formal name? Thanks in advance!

Anixx
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Steve Beresh
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    Can you check I TeXed the equation correctly? (A consequence of the timing of the edit is that your example disappeared - it looked like was just to clarify what the equation was supposed to be, so I guess that isn't a problem, but either of us can put it back if you still want it). – mdp Apr 24 '13 at 17:06
  • You have it right; thanks for the edit! :) – Steve Beresh Apr 24 '13 at 17:12

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Almost certainly not! There are infinitely many identities like this.

The reason this works is that the two terms are $\left(1-\frac{1}{x}\right)^{-x} \times (x-1)$ and $\left(1-\frac{1}{x-1}\right)^{-(x-1)} \times (x-2)$, where in both cases the first term $\to e$ as $x\to\infty$ by a statement closely related one of the standard definitions of $e$.

Thus this is more a corollary of the statement $\lim_{n\to\infty}(1+1/n)^n = e$ than an independent fact.

not all wrong
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  • I can see how $\left(1-\frac{1}{x}\right)^{-x}$ and $\left(1-\frac{1}{x-1}\right)^{-(x-1)}$ are related to $\lim_{n\to\infty}(1+1/n)^n = e$, but where do the $\times (x-1)$ and $\times (x-2)$ fit in with the standard definition?

    I suppose the reason I found the relationship intriguing is because when you graph the standard definition, on the positive side of the x-axis the line comes from the bottom, comes up, and evens out at y=e. When you graph my relationship, on the positive x-axis the line comes from up top rather than the bottom.

     – Steve Beresh Apr 24 '13 at 18:04
  • (a) It's simply that $e\times(x-1)-e\times(x-2) = e$, or $(x-1)-(x-2) = 1$... (b) The same is true of $\left(1-\frac 1 x \right)^{-x}$; it decreases to $e$ from above! – not all wrong Apr 24 '13 at 23:55
  • Ah, thanks. Makes sense now. Much obliged! – Steve Beresh Apr 25 '13 at 01:57
  • No problem! I feel slightly guilty for cheating slightly to make my point clearer though - there are $1/x$ terms in the $e$ part, but they're the same so they cancel. – not all wrong Apr 25 '13 at 09:01