The limit of limn→∞exp(−1+exp(−2+exp(−3+⋯exp(−n)⋯)))\lim\limits_{n\to\infty} \exp(-1+\exp(-2+\exp(-3+\cdots\exp(-n) \cdots))).

Does the following limit exist ?

limnexp(1+exp(2+exp(3+exp(n))))

If yes, can it be expressed in a closed form ?

PARI shows the following numerical value :

n=-100;x=exp(n);while(n<-1,n=n+1;x=exp(x+n));print(x)

0.4241685586940448516119410516

Within this precision, 200 yields the same value.

Answer

The limit exists because the sequence is monotonic and bounded, although I don’t know yet what the limit is. Let’s denote this sequence En.

The sequence is monotonic because n+exp((n+1))>n, therefore exp(n)+exp((n+1)))>exp(n), therefore exp((n1)+exp(n)+exp((n+1))))>exp((n1)+exp(n)), etc, until you get En+1>En.

The sequence is bounded because n+exp(negative number)<0, therefore En<exp(0)=1.

By monotone convergence theorem En converges.

Attribution
Source : Link , Question Author : Peter , Answer Author : Michael

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