$e^{x} > 1$ and $0 < e^{x} < 1$

Question

So $$\exp(x) := \sum_{n=0}^{\infty} \frac {x^n} {n!}$$

How to prove that $\exp(x) > 1$ when $x > 0$ and moreover $\exp(x) < 1$ when $x<0$

Is it possible with induction? Or must I use Cauchy product?

Answer 1

The power series of the exponential function is defined on $\Bbb R$ so we can differentiate it term by term on $\Bbb R$ and we get

$$\exp'(x)=\exp(x)$$

Moreover, we see easily that $\exp(x)>0$ for $x\ge0$ and using the Cauchy product we get $$\exp(x)\exp(y)=\exp(x+y),\quad \forall (x,y)\in\Bbb R^2$$ hence

$$\exp(-x)\exp(x)=\exp(0)=1\implies \exp(x)>0,\; \forall x<0$$

hence $\exp$ is strictly increasing function on $\Bbb R$ and then

$$\exp(x)>\exp(0)=1,\quad \forall x>0$$

$$\exp(x)<\exp(0)=1,\quad \forall x<0$$

Answer 2

For $x>0$, $$\exp(x)=1+x+\frac{x}{2}+\cdots>1+0+0+\cdots=1$$

Since $\exp(0)=1$ and $\exp$ is strictly monontone increasing, $\exp(x)<1$ for $x<0$.

Answer 3

If $x>0$, you have a series with positive terms, so its sum is greater than the sum of the first two terms, which is $1+x$ and $1+x>1$.
If $x<0$ you have an alternating series. It's a theorem that for alternating series, the error bound when you take the sum up to rank $m$: $\sum_{k=0}^m a_k$, is at most $\lvert a_{m+1}\rvert$ and the error has the sign of $a_{m+1}$. So here, if you take the sum up to rank $0$ as an approximation of $\mathrm e^x$, the error is negative, i.e. $\mathrm e^x<1$.

Answer 4

the first part $e^x > 1$ for $x > 0$ follows from $e^x = 1 + x + \frac{x^2}{2} + \cdots$ because $x > 0$ implies $x^2 > 0, x^3 > 0, \cdots$

the second part $0 < e^x < 1$ for $x < 0$ does not seem to be elementary. if you can establish even an instance of the additive property $e^{-x} = \dfrac{1}{e^x},$ then you are done.

one way of establishing the additive property is to claim the $e^x$ is the unique solution of $\frac{dy}{dx} = y, y = 1 \text{ at } x = 0.$

Answer 5

It was already mentioned in comments and other answers that the inequality $\exp(x)>1$ for $x>0$ is easy.

Here is rather elementary proof that $\exp(x)\le1$ for $x<0$.

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Let us denote $$f_n(x)=1+x+\frac{x^2}{2!}+\dots+\frac{x^n}{n!}.$$

Notice that $f_{n+1}'(x)=f_n(x)$.

Try to prove by induction that for $n=1,2,\dots$ that:

• $f_{2n-1}(x)$ is increasing and it has only one real root;
• $f_{2n}(x)>0$ for each $x\in\mathbb R$.

(In the inductive step you can use that fact that $f_{2n}(x)$ has only one critical point $c$, which is exactly the root of $f_{2n-1}(x)$. At this point $c$ we have $f_{2n}(c)=f_{2n-1}(c)+\frac{c^{2n}}{(2n)!}=\frac{c^{2n}}{(2n)!}>0$.)

Now since each $f_{2n-1}(x)$ is increasing, we get $f_{2n-1}(x)<f_{2n-1}(0)=1$ for $x<0$. Therefore we have for $\exp(x)=\lim\limits_{n\to\infty} f_{2n-1}(x)$ that $$\exp(x)\le1$$ for $x<0$.

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So now the only problem is how to show that the inequality is strict. I will try the following approach, maybe someone can see a more elegant argument.

We already know that the function $\exp(x)$ is non-decreasing. (As a pointwise limit of a sequence of increasing functions.)

So if $\exp(x)=1$ for some $x<0$, then this function would be constant on the interval $[x,0]$. So it suffices to show that we can find values smaller than $1$ close to zero.

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Now we notice that we have $$\exp(-1) = 1-1+\frac1{2!}+\frac1{4!}\dots = \frac{1}{2!}+\frac1{4!}+\dots \le -\frac1{2}+\frac1{2^2}+\frac1{2^3}+\dots =1.$$

So we have $\exp(-1)\le-1$. (And the above argument could be relatively easily modified to get $\exp(-1)<-1$.)

Let us try whether we can use similar argument for $-1<x<0$. In this case we get

$$\exp(x) \le 1+x + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots \le 1+x + \frac{x^2}{2!} + \frac{x^2}{4!} + \dots =1+x+x^2\left(\frac1{2!}+\frac1{4!}+\dots\right) \le 1+x+x^2\left(\frac12+\frac1{2^2}+\dots\right) = 1+x+\frac{x^2}2 = \frac{1+(1+x)^2}2.$$

Since the function $\frac{1+(1-x)^2}2$ has values smaller than $1$ on the interval $(-1,0)$, the same is true for $\exp(x)$.

Thus $\exp(x)$ cannot be constant on any interval of the form $[x,0]$ and we get, that the inequality is strict.