Function that describes the approach of (1 + 1/n)^n to Euler's number. What is the limit? I will try to state the function in simple parts. First, take 
$\lim\limits_{n \to ∞} (1+\frac{1}{n})^n$, then subtract e from each value of n. Make the reciprocal of this function. What is the limit of the derivative of this function?
$ \frac{d}{dn}\lim\limits_{n \to ∞} 1/(e-(1+\frac{1}{n})^n)$,
One way of defining Euler's number is the limit as n approaches infinity of (1 + 1/n)^n. I have used Excel to compute each value up to n = 1,000. I then calculated the inverse of the difference of e for each value of n. For n= 1, we find (1+1/1)^2=2. Then, 1/(e - 2) = 1.3922... For n = 2, we have (1+1/2)^2= 2.25. Then, (1/e-2.25) = 2.1355... 
This value increases as n increases, as expected. I then calculated the difference between the value and the one before it. I found that this value decreases to a point, but does not do so smoothly. When graphed after about n = 250 there is an erratic but downward trend. However, I'm not sure if this erratic behaviour is due to successive rounding errors by Excel.The average value from n = 900 to n= 1000 is 0.735758909. The series appears to converge very quickly, at just n = 5 we have 0.736817183. What is exact value of this limit? Is it related to any important constants?
 A: Basically, what you need is to quantify in a precise sense "at which rate $\left(1+\frac{1}{n}\right)^n$ does approach its limit $e$."
To do so, one of the preferred and very useful methods is to use Taylor expansions$^{(\dagger)}$
$$\begin{align}
\left(1+\frac{1}{n}\right)^n 
&= e^{n\ln\left(1+\frac{1}{n}\right)}
= e^{n\left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)}
= e^{1-\frac{1}{2n}+o\left(\frac{1}{n}\right)}
= e\cdot e^{-\frac{1}{2n}+o\left(\frac{1}{n}\right)}
\\&= e\cdot \left(1-\frac{1}{2n}+o\left(\frac{1}{n}\right)\right)
= e - \frac{e}{2n}+o\left(\frac{1}{n}\right)
\end{align}$$
so we get
$$ \left(1+\frac{1}{n}\right)^n  - e \operatorname*{\sim}_{n\to\infty} -\frac{e}{2n}.$$
(Which means, in a formal sense, that "the difference behaves like $-\frac{e}{2n}$ when $n$ tends to $\infty$")
In particular, what you appear to be considering (the way you wrote it is slightly off, the derivative should not be outside the limit) is 
$$
\frac{1}{n}\cdot \frac{1}{e-\left(1+\frac{1}{n}\right)^n} \xrightarrow[n\to\infty]{}\frac{2}{e}.
$$
Note that $\frac{2}{e} \simeq 0.73575888$, so this is indeed consistent with your numerical experiments.

$(\dagger)$ Namely, two standard ones, to low order: $\ln(1+x) = x-\frac{x^2}{2} + o(x^2)$ and $e^x = 1+x + o(x)$ when $x\to 0$.
