Euler Euler?

and hence $\Omega = e^{e^2} \approx 1618.177$ .

$b_{n}$ is a sequence which tends to $e$ as $n \to \infty$ . We can use Stirling's Approximation

as $\large b_{n} \to lim_{n\to\infty} e(n+1) - ne = e$

So $\large (1+\frac{b_{n}}{n})$ can be approximated quite well to $\large 1+\frac{e}{n}$ as $n\to\infty$ .

Next for $\large \sum_{r=0}^{n} \frac{n^{r}}{r+1}\binom{n}{r} = \frac{(n+1)^{n}}{n}-\frac{1}{(n+1)(n)}$

You can use integration of $(1+x)^n$ from $0$ to $n$ and the well known identity of binomial theorem to prove this . Or you can use basic rearrangement of the series. This is a well known result...hence I am not proving this.

So a good approximation for this limit would be:-

$\large \lim_{n\to\infty}(1+\frac{e}{n})^{{\frac{1}{n^{n-2}}(\frac{(n+1)^{n}}{n}-\frac{1}{(n+1)(n)}})}$

So now using the formula for $1^{\infty}$ form of limits we have our approximation as :-

$\Large e^{\lim_{n\to\infty}(\frac{e}{n}\frac{n^{2}}{n^{n}}(\frac{{n+1}^{n}}{n} - \frac{1}{(n+1)(n)})}$

$\Large e^{\lim_{n\to\infty}(\frac{e(n^{2})}{n}(\frac{{(1+\frac{1}{n})}^{n}}{n} - \frac{1}{(n^{n})(n+1)(n)})}$

Using the fact that $(1+\frac{1}{n})^{n}$ tends to $e$ and $\frac{1}{n^{2}}(\frac{1}{n^{n}(n+1)})$ tends to 0 as $n\to\infty$ we have another approximation as :-

our limit is approximated to

$\Large e^{\lim_{n\to\infty}\frac{e\cdot e\cdot n^{2}}{n^{2}}}$

$\large =e^{e\cdot e}$

$\large e^{e^{2}}$ .

now as for the approximations I do not have any formal proofs for those approximations....but I post the pictures of the graphs....perhaps someone more skilled than me at real analysis could prove these.