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Let $a_1=a\neq1$ and $a_{n+1}=\dfrac{1}{S_n-1}+1$ then find the $\lim_{n\to \infty}a_n=?$

such that :

$$S_n=\sum_{i=1}^n a_i$$

$$a_1=a \\a_2=\dfrac{1}{a-1}+1 \\a_3=2a+\dfrac{1}{a-1}+1 \\ a_4=3a+2+\dfrac{2}{a-1}+\dfrac{a-1}{2a^2-2a+1} $$

now what do I do ?

Especially Lime
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Almot1960
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    This is at least the third installment of exactly the same question... Why the pollution? – Did Jan 22 '18 at 08:32
  • @Did I don't think this should be considered a duplicate of the linked question. While the two problems turn out to be equivalent, that is far from obvious from the quite different definitions of the sequence. – Especially Lime Jan 22 '18 at 10:34
  • @EspeciallyLime Sorry but I beg to differ. 1. The equivalence was already explained to the OP in at least one of their previous installments of this question, now deleted. 2. They were probably aware of the equivalence even before it was formally explained to them since they started to fire repeatedly this question just after the other one was posted. 3. Posting, receiving explanations, deleting, reposting, and so on ad lib, is a big no-no for this site, and it should be, don't you think? – Did Jan 22 '18 at 16:22
  • @Did Yes, what you describe should be a big no-no. But the linked question wasn't posted by the same person and seems to me genuinely different. Perhaps there are good reasons to close this question, but IMO the original justification given for closing it as a duplicate is not a good one. – Especially Lime Jan 22 '18 at 17:35
  • @EspeciallyLime Once again, this is at least the third installment by the same OP of the same question (which they delete afterwards). Furthermore, it was already explained that it is mathematically equivalent to the question asked earlier by another user. Thus, "duplicate" seems a quite apt description here (to which one could even add "deliberate"). – Did Jan 22 '18 at 18:08
  • @Almot Which parts of this question are not already fully addressed on the other page? (One could also wonder about the psychological motivation of reposting exactly some question that just got solved, but let us leave this mystery aside.) – Did Jan 23 '18 at 23:12

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