Proving that if $S_n$ is monotone increasing and unbounded, then $\exp(-S_n)$ converges?

Question

Prove that if $S_n$ is monotone increasing and unbounded, then $\exp(-S_n)$ converges as $n \to +\infty$

I'm guessing it converges to $0$ but I can't prove it. Since $S_n$ monotonically increasing can we say something like $-S_n \le -S_1$, i.e. $e^{-S_n} \le e^{S_1}$? I'm supposed to argue from definitions. The problem is I usually work backwards for the $\epsilon$ to find a suitable $N$, but for this problem that doesn't seem viable.

Answer 1

$S_n$ must be tending to $+\infty$, since it is unbounded and increasing monotonically.

This means we can pick any $M>0$ and eventually, there will be an $N\in \mathbb{N}$ with $S_n>M$ for any $n>N$.

Then, for any $\epsilon>0$, we can take $N$ big enough such that $$ S_n>\log\frac{1}{\epsilon}\implies-S_n<\log\epsilon\implies e^{-S_n}<\epsilon $$

Answer 2

We can assume that $S_n>0$ for all $n$. From $e^{S_n}=1+S_n+ \frac{S_n^2}{2!}+...> S_n$ we get

$0 \le e^{-S_n}<\frac{1}{S_n}$.

Can you proceed ?

Answer 3

1) $S_{n+1} \ge S_n$ , increasing.

2) Unbounded:

For $M >0$, real, there exists a $n_0$ such that for $n\ge n_0$ we have $S_n >M$.

Consider: $a_n:= \exp(-S_n)= \dfrac{1}{\exp (S_n)}$.

Note : $\exp$ is an increasing function:

Hence:

$a_{n}= \dfrac{1}{\exp(S_n)} \ge \dfrac{1}{\exp(S_{n+1})}= $

$a_{n+1},$ i.e.

$a_n$ is a decreasing sequence.

Since $a_n \gt 0$ for $n \in \mathbb{Z^+},$

$0$ Is lower bound.

Monotonically decreasing sequence a_n bounded below:

Limit exists.

Note: We have not used property 2).

Where does it come in?

Answer 4

Since $\exp(-x)$ is monotonously decreasing, and $S_n$ is monotonously increasing, $\exp(-S_n)$ is monotonously decreasing. On the other hand, $\exp(-S_n)>0$ for all $n$, thus the sequence is lower bounded by $0$.

A monotonously decreasing and lower-bounded sequence converges.