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Does $a_n$ converge if$a_{n+1} = a_n + \frac{1}{e^{a_n}+1}$? Or does its convergence depend on $a_0$? As described in the title, it seems intuitively that it should converge, but I don't know how to prove it.

J. W. Tanner
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n yk
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  • it diverges for any value of a_0 – Mittens Feb 05 '24 at 04:40
  • To see more intuitively why this diverges, replace the equation $\Delta a_n =a_{n+1}-a_n=1/(e^{a_n}+1)$ with its continuous analog, $y'=1/(e^y+1)$. You can see that the derivative exists and is positive everywhere, so this function cannot converge to any finite limit, since the derivative there would be $0$ or would not exist. – Alexander Burstein Feb 05 '24 at 04:45
  • Such a unique idea, thank you. – n yk Feb 05 '24 at 05:23

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$a_n$ is nondecreasing. If $a_n$ converges then the series $a_{n+1} -a_n \to 0$ converges and so $\lim\limits_{n\to \infty} \frac1{e^{a_n} + 1} =0$ so $\frac1{e^{\ell} + 1} = 0$ which is impossible. This implies that $\lim\limits_{n\to\infty} a_n = \infty$

Kroki
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