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This constant is half of the Silver ratio, $1+\sqrt{2}$, which is analogous to the golden ratio. European paper (A4, A5, etc.) is sized by the silver ratio, if you remove the largest square.


Both sequences have $e$ as the limit, so it suffices to show that the left sequence is increasing and the right sequence is decreasing. Use the AM-GM inequality to get $$\frac{n+2}{n+1}=\frac{\frac{n+1}n+\ldots+\frac{n+1}n+1}{n+1}>\sqrt[n+1\,]{\left(\frac{n+1}n\right)^n}$$ It follows that $$\left(1+\frac1n\right)^n<\left(1+\frac1{n+1}\right)^{n+1}$$ ...


The sequence $\left(1+\frac{1}{n}\right)^n$ is strictly increasing and converging to $e$. This implies the first inequality. $\left(1+\frac{1}{n}\right)^{n+1}$ is strictly decreasing and converging to $e$. This implies the first inequality.


You can prove the upper bound directly without resorting to the full binomial expansion or the Taylor series. Let $b_n = \left(1+\frac{1}{n}\right)^{n+1}$. Claim $b_n$ is decreasing. Indeed, \begin{align*} \frac{b_{n+1}}{b_n} &= \left(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\right)^{n+1}\left(1+\frac{1}{n+1}\right) \\ &= ...


Hint: note that $$\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e$$


You already expanded it, so look at the individual terms: $$\binom nk\frac1{n^k}=\frac{n(n{-}1)\ldots(n{-}k{+}1)}{k!}\cdot\frac1{n^k}<\frac{n^k}{k!}\cdot\frac1{n^k}=\frac{1}{k!}$$ Therefore $$\biggl(1+\frac1n\biggr)^n<\sum_{k=0}^n\frac1{k!}<2+\sum_{k=2}^\infty\frac1{k!}<2+\sum_{k=1}^\infty\frac1{2^k}=2+1=3$$


Let $X$ be the number of missing pulses. As you wrote, $\Pr(X=0)=e^{-0.2}$. That is the answer, no further manipulation needed. If $Y$ is the number of missing pulses on $2$ disks, then $Y$ has Poisson distribution with parameter $0.4$. So the probability of no missing pulses is $e^{-0.4}$. For this particular problem, we could instead square the first ...

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