Proof: exp(-1/x) < 1 for x > 0 (real numbers)

Question

How can i proof the following inequality:

exp(-1/x) < 1 for x > 0 (real numbers)

First i tried to assume that exp(-1/x) is equal to sum ((-1/x)^n/n!) from n=0 to infinite < 1 but i cant simplify the inequality further. Do you have any tips or idea to proof this?

Answer 1

$$ \exp(t) = 1 + t + \frac{t^2}{2!} + \frac{t^3}{3!} + \ldots $$ so $$ \exp(\frac{-1}{x}) = 1 - \frac{1}{x} + \frac{1}{2!x^2} - \frac{1}{3! x^3} \pm \ldots, $$ which is an alternating series (for $x > 1$) of nonzero terms, whose sum is therefore strictly less than its first term.

N.B.: This only works for $x > 1$, and therefore does not prove the claim for all $x > 0$; that appears to require a good deal more work.

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As Barry Cipra points out, I've omitted something important: the terms need to be decreasing for my alternating series claim to be correct. For $x > 1$, this is clear. But what about for $0 < x < 1$? In the case, my argument fails. So this is only a partial answer, and I suggest OP un-accept it, since it's incomplete.

Answer 2

You just need to know that the exponential function $t \mapsto e^t $ is strictly increasing at $\mathbb R $ and of course $e^0=1$.

from this

$$x>0\implies 1/x>0$$ $$\implies -1/x <0 \implies e^{-1/x}<e^0$$

Answer 3

To prove $e^{-\frac 1 x}<1$, it is sufficient to prove $e^{t}<1 ~\forall t<0$

Since the function $f(t)=e^t$ is monotonously increasing $(f'(t)=e^t>0)$, and $f(0)=1$, we can conclude that $f(t)<f(0) ~\forall t<0$. That is nothing else than saying :

$$e^t<1~ \forall t\in \mathbb{R^-} \leftrightarrow e^{-\frac{1}{x}}<1 ~\forall x>0$$

Answer 4

So as not to be misled by the exponential notation, suppose all we have to start with is a function $f$ given by the power series

$$f(u)=\sum_{n=0}^\infty{u^n\over n!}=1+u+{u^2\over2}+{u^3\over6}+\cdots$$

It's clear that $f(0)=1$ and $f(u)\gt1$ if $u\gt0$. For this problem, however, setting $u=1/x$ with $x\gt0$, we need to show that

$$f(-u)=\sum_{n=0}^\infty(-1)^n{u^n\over n!}=1-u+{u^2\over2}-{u^3\over6}+\cdots\lt1\quad\text{for }u\gt0$$

The easiest way to do this (I think) is to show that $f(-u)=1/f(u)$ and then invoke the fact that $f(u)\gt1$ for $u\gt0$, and the easiest way to do this (I think!) is to establish the general property

$$f(u+v)=f(u)f(v)$$

and then invoke the fact that $f(u-u)=f(0)=1$. So let's do that:

$$\begin{align} f(u+v) &=\sum_{N=0}^\infty{(u+v)^N\over N!}\\ &=\sum_{N=0}^\infty{1\over N!}\sum_{n=0}^N{N\choose n}u^nv^{N-n}\\ &=\sum_{N=0}^\infty{1\over N!}\sum_{n=0}^N{N!\over n!(N-n)!}u^nv^{N-n}\\ &=\sum_{N=0}^\infty\sum_{n=0}^N\left({u^n\over n!}\cdot{v^{N-n}\over(N-n)!}\right)\\ &=\sum_{n=0}^\infty\sum_{N=n}^\infty\left({u^n\over n!}\cdot{v^{N-n}\over(N-n)!}\right)\qquad(*)\\ &=\sum_{n=0}^\infty\left({u^n\over n!}\sum_{N=n}^\infty{v^{N-n}\over(N-n)!} \right)\\ &=\sum_{n=0}^\infty\left({u^n\over n!}\sum_{m=0}^\infty{v^m\over m!} \right)\\ &=\left(\sum_{n=0}^\infty{u^n\over n!}\right)\left(\sum_{m=0}^\infty{v^m\over m!}\right)=f(u)f(v) \end{align}$$

The key step, $(*)$, is the switching of the order of summation. This technically requires its own justification, which amounts to showing (or knowing) that the series is absolutely convergent.

Remarks: For $0\lt u\lt1$ it's clear from the strictly decreasing nature of the terms of the alternating series $f(-u)=1-u+{u^2\over2}-{u^3\over6}+\cdots$ that $0\lt f(-u)\lt1$ without any of the preceding folderol. But for $u\gt1$ it's not at all clear just from the series definition that $f(-u)$ is either positive or less than $1$. E.g., it's really hard to see, just from its opening, where the sum

$$1-10+50-{500\over3}+{1250\over3}-\cdots$$

is going to settle down. If there is an easier way to prove that $f(-u)\lt1$ for $u\gt0$ than the approach I've taken, I'd very much like to see it.