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I am stuck with this equation

Prove $$\lim_{n\to\infty} \frac{1-(1-p)^n -np(1-p)^{n-1}}{1-np(1-p)^{n-1}} \text{ equal to 0.4180 where } p=\frac{1}{n}$$

So here what I did: divide by ${(1-p)^{n-1}}$

$$ \frac{\frac{1}{(1-p)^{n-1}} - \frac{(1-p)^n}{(1-p)^{n-1}} - \frac{np(1-p)^{n-1}}{(1-p)^{n-1}}} {\frac{1}{(1-p)^{n-1}} - \frac{np(1-p)^{n-1}}{(1-p)^{n-1}}}$$ $$ $$ $$\frac{\frac{1}{(1-p)^{n-1}} - (1-p) -np} {\frac{1}{(1-p)^{n-1}} - np } $$ $$$$ $$ \frac{\frac{1}{(1-p)^{n-1}}-np} {\frac{1}{(1-p)^{n-1}} - np } - \frac{(1-p)} {\frac{1}{(1-p)^{n-1}} - np }$$ $$$$ $$ 1 - \frac{(1-p)} {\frac{1}{(1-p)^{n-1}} - np }$$

$\text{substitute } p=\frac{1}{n} $

$$ 1 - \frac{(1-\frac{1}{n})} {\frac{1}{(1-\frac{1}{n})^{n-1}} - n \times \frac{1}{n} }$$

$$ 1 - \frac{(1-\frac{1}{n})} {\frac{1}{(1-\frac{1}{n})^{n-1}} - 1 }$$

After this, I tried several things but never get the required result!

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  • $\begingroup$ I just noticed if I substitute n=100, 1000,.... I get the desirable result $\endgroup$ – SHADOW.NET Jan 24 '17 at 9:42
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$$ \begin{align} \color{red}{L} &=\lim_{n\rightarrow\infty}\frac{1-(1-p)^n-np(1-p)^{n-1}}{1-np(1-p)^{n-1}} \\[2mm] &=\lim_{n\rightarrow\infty}\left[\frac{1-np(1-p)^{n-1}}{1-np(1-p)^{n-1}}-\frac{(1-p)^n}{1-np(1-p)^{n-1}}\right] \qquad\qquad{\small\{p=1/n\}} \\[2mm] &=\lim_{n\rightarrow\infty}\left[1-\frac{(1-1/n)^n}{1-(1-1/n)^{n-1}}\right] \space\qquad\qquad\qquad{\small\{1/e=\lim_{n\rightarrow\infty}(1-1/n)^n\}} \\[2mm] &=1-\frac{1/e}{1-1/e} =\color{red}{1-\frac{1}{e-1}} \,\,\approx 0.41802329313067357561499799489099\dots \\[2mm] \end{align} $$

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