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I was observing here and conjecture $(1)$

$$\lim_{n\to\infty}{S_{n-1}S_{n+2}\over S_nS_{n+1}}=e^2\tag1$$

Given that Harlan Brothers' formula $(2)$

$$\lim_{n\to\infty}{S_{n-1}S_{n+1}\over S_n^2}=e\tag2$$

Trying to prove $(1)$:

$(1)\div(2)$

$$\lim_{n\to\infty}{S_{n-1}S_{n+2}\over S_nS_{n+1}}\times{S_n^2\over S_{n-1}S_{n+1}}=e\tag3$$

Simplified to

$$\lim_{n\to\infty}{S_{n}S_{n+2}\over S_{n+1}^2}=e\tag4$$

$(4)$ which is the same as $(2)$, so hence we can say $(1)$ is correct?

2nd conjecture

Another conjecture from observing $(1)\div(2)^2$, simplified to

$$\lim_{n\to\infty}{S_n^3S_{n+2}\over S_{n-1}S_{n+1}^3}=1\tag5$$

I noticed that it takes the binomial coefficients,you can see a pattern emerges

$$\lim_{n\to\infty}{S_n^4S_{n+2}^4\over S_{n-1}S_{n+1}^6S_{n+3}}=1\tag6$$

$$\lim_{n\to\infty}{S_n^5S_{n+2}^{10}S_{n+4}\over S_{n-1}S_{n+1}^{10}S_{n+3}^5}=1\tag7$$

and so on ...

We can write as

$$\lim_{n\to \infty}\prod_{k=0}^{m}S_{n+k-1}^{(-1)^{k+1}{m\choose k}}=1\tag8$$ $m\ge3$

Numerically we have verified for certain range of $S_n$, but it is not necessarily indicate that it is true for larger values of $S_n$

How can we prove $(8)?$

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The first claim is correct, but a better way to put it would be this:

  1. Harlan Brothers says: $\lim_{n\to\infty}{S_{n-1}S_{n+1}\over S_n^2}=e\tag2$

  2. Index-shift to get $\lim_{n\to\infty}{S_{n}S_{n+2}\over S_{n+1} ^2}=e\tag2$

  3. Limit of product is product of limits if both exist, hence

  4. \begin{align} e^2 &= \lim_{n\to\infty}{S_{n}S_{n+2}\over S_{n+1} ^2}\lim_{n\to\infty}{S_{n-1}S_{n+1}\over S_n^2}\\ &= \lim_{n\to\infty}{S_{n}S_{n+2}\over S_{n+1} ^2}{S_{n-1}S_{n+1}\over S_n^2}\\ &= \lim_{n\to\infty}{S_{n+2}\over S_{n+1} }{S_{n-1}\over S_n} \end{align}

This way, you're not working from the thing you're trying to prove, but rather from things you know to be true.

As for the second one...I didn't really look, because you should, in general, ask one question at a time. But it sure looks like a not-too tough induction proof.

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  • $\begingroup$ By the way: thanks for introducing me to the Harlan Brothers' formula -- I'd never seen it before. $\endgroup$ – John Hughes Feb 2 '17 at 12:50
  • $\begingroup$ Thanks for providing a nice proof and I will try and take your advice, one question at a time. $\endgroup$ – gymbvghjkgkjkhgfkl Feb 2 '17 at 12:53

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