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As the title states, I'm trying to prove $a_{n+1}=1-e^{-a_n}$ has no infimum if $a_1<0$. I'm not sure this is true, but calculating $a_n$ for large $n$ leads me to believe it is. I've already proved $a_{n}$ is monotone decreasing, and that if $a_{1}>=0$ then its infimum is 0.

I know the very basics of series and sequences; no Taylor and the such, but most limit calculation theorems as well as a decent amount (L'hopital, integrals, etc...) for calculating limits in $R$.

Any hints would be appreciated!

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I posted too hastily. Solution: assume the sequence is bounded from below, then, since it is monotone decreasing (when $a_1!=0$), it has a limit. We know that $a_1<0$ which means $a_n<0$ for every n. But assuming the existence of the limit, and the fact that the limits $a_n$ and $a_{n+1}$ are equal when $n->\infty$, we can compare the two and come up with the fact that the limit can only be $0$ - which is a contradiction. – rowe Nov 14 '11 at 18:11
(I will either post this as an answer in 8 hours or accept the first answer I like before that) – rowe Nov 14 '11 at 18:12
Correction: meant to type 'a finite limit' there. – rowe Nov 14 '11 at 18:18
For me, it was easier to first let $b_n=-a_n$. But I may be unusually sensitive to minus signs. – André Nicolas Nov 14 '11 at 18:22
up vote 1 down vote accepted

I will assume that the sequence is real-valued, $a_n \in \mathbb{R}$.

Since $1+x \le \exp(x)$ it follows, by using $x=-a_n$, that $1 - a_n \le \exp(-a_n) = 1-a_{n+1}$, thus the series is non-increasing $a_{n+1} \le a_{n}$.

Suppose the sequence converges to $a^\ast$, i.e. $\lim_{n \to \infty} a_n = a^\ast$. Then $a^\ast = 1-\exp(-a^\ast)$. The only finite real solution to this equation is $a^\ast = 0$. Another, infinite one, is $a^\ast = -\infty$.

Since the sequence is non-increasing, it follows that $a_{n} \le a_{1}$. If $a_1 < 0$, all elements are negative, $\vert a_n \vert \ge \vert a_1 \vert > 0$, hence $a_n$ can not converge to 0, so it follows that the sequence with $a_1 < 0$ must diverge to $-\infty$.

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