Proving positivity of the exponential function Question. Without using the semigroup property ($\mathrm{e}^{x}\mathrm{e}^{y}=\mathrm{e}^{x+y}$),
how can we show that $\mathrm{e}^{x}>0$ for all $x\in\mathbb{R}$ only by using the series expansion?
Explanation.
From the series expansion of $\mathrm{e}^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$ for $x\in\mathbb{R}$, we see that $\mathrm{e}^{x}>0$ for $x\geq0$.
Thus, if the series becomes negative, this can only happen for negative values of $x$.
So proving $\mathrm{e}^{-x}$ for $x>0$ will complete the proof.
As the series converges uniformly on any compact interval $I\subset\mathrm{R}$, we can rearrange the terms of the series and write
$\mathrm{e}^{-x}=\lim_{n\to\infty}g_{n}(x)$ for $x\geq0$, where $g_{n}(x):=1+\sum_{k=1}^{n}\Big(\frac{x^{2k}}{(2k)!}-\frac{x^{2k-1}}{(2k-1)!}\Big)$ for $x\geq0$ and $n\in\mathbb{N}$.
Obviously, $g_{n}$ is decreasing on $[0,1]$ and $g_{n}(1)>\frac{1}{\mathrm{e}}$.
I need to prove the following.
Claim. There exists an increasing divergent sequence $\{\xi_{n}\}\subset(0,\infty)$ such that $g_{n}$ is decreasing on $[0,\xi_{n}]$ with $g_{n}(\xi_{n})>0$ for $n\in\mathbb{N}$.
Strengthened Claim. $\xi_{n}:=\sum_{k=1}^{n}\frac{1}{k}$ for $k\in\mathbb{N}$.
 A: A hyperbolic trigonometry approach. Set
$$
C(x)=\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\quad\text{and}\quad S(x)=\sum_{k=1}^\infty\frac{x^{2k-1}}{(2k-1)!}
$$
It suffices to show that $C(x)>S(x)$, for every $x\in\mathbb R$.
First observe that: $C'(x)=S(x)$ and $S'(x)=C(x)$. Then observe that
$$
\big(C^2(x)-S^2(x)\big)'=2\big(C(x)C'(x)-S(x)S'(x)\big)=2\big(C(x)S(x)-S(x)C(x)\big)=0,
$$
and hence 
$$
C^2(x)-S^2(x)=C^2(0)-S^2(0)=1.
$$
Thus, for every $x\in\mathbb R$,
$$
C(x)=\sqrt{S^2(x)+1}>S(x).
$$
A: Using termwise differentiation one finds that $\exp$ satisfies the linear differential equation $y'=y$, which obvioulsy satisfies the assumptions of the existence and uniqueness theorem. The function $y_0(x):\equiv0$ is a solution, and no other solution can cross the graph of $y_0$. It follows that $x\mapsto e^x$, which is positive when $x=0$, is positive on its full domain ${\mathbb R}$.
A: Assuming uniform convergence of the series you can show by termwise differentiation that $f(x) = e^x$ verifies $f'(x) = f(x).$ 
Clearly $e^x = \sum_k \frac{x^k}{k!}$ is strictly positive for all positive $x$ therefore it is an increasing function on $\mathbb{R}^+$. Consider the set $A = \{x < 0 : e^x \leq 0 \}$ and assume that it is non empty. 
Let $(x_n)_{n\in \mathbb{N}}$ be a sequence in $A$ that converges to $L$. Then $L \in A$ by continuity of $f$
$$ f(L) = f(\lim_{n \to \infty} x_n) =  \lim_{n \to \infty} f(x_n) \leq 0.$$
Therefore $a := \sup A \in A$ and $a < 0$ and $e^a \leq 0$.
If $e^a < 0$ then notice that $e^0 = 1$ and by the intermediate value theorem there exists $ a < c < 0$ such that $e^c = 0$ and $c \in A$ which contradicts the maximality of $a$.
If $e^a = 0$ then consider $$C = \{c' \leq a \; \vert \forall x \in (c',a], \; \;f(x) = 0 \}.$$ If $\inf C = k > - \infty$, then $\exists \delta > 0$ s.t. the interval $[k- \delta, k + \delta]$ around $k$ is such that $f > 0$ on $[k- \delta, k)$ or $f < 0$ on $[k- \delta,k)$.
Since $f$ is equal to its own derivative it is either positive-increasing or negative increasing on $(k-\delta,k$). In both cases by the mean value theorem, $\exists \alpha \in (k-\delta/2,k)$ s.t. 
$$ f'(\alpha) = \frac{f(k) - f(k- \delta/2)}{\delta/2} = - \frac{f(k-\delta/2)}{\delta/2} $$
This is a contradiction since $f'(\alpha) = f(\alpha)$ and $f(k- \delta/2)$ have the same sign and are both not equal to zero.
If $\inf C' = - \infty$ then $e^x = 0 \forall x \leq a.$ We must proceed differently:
Consider the function $F : \mathbb{R}^+ \rightarrow \mathbb{R}: x \mapsto F(x) = \int_{a+x}^0 e^t \;dt$
Clearly $$F(x) = [e^t]_{a+x}^0 = 1 - e^{a+x}.$$
Using the change of variable $u(t) = t - x $ in the integral we get 
$$ F(x) = \int_{x+a}^0 e^t dt = \int_{a}^{-x} e^t dt = [e^t]^{-x}_a = e^{-x} - e^a = e^{-x}$$
Therefore $\forall x > 0$:
$$ 1 - e^{a + x} = e^{-x}.$$
Since $e^t > 0 \; \forall t > a$ the exponential function is increasing on $(a, + \infty)$  so 
$$a + x > 0 \Rightarrow  e^{a + x} > e^{0} = 1 \iff F(x) = 1 - e^{a+x} < 0.$$ 
But $ a + x > 0 \Rightarrow -x < a$ and $e^{-x} = 0$ so $$F(x) = e^{-x} = 0$$
which is a contradiction.
We conclude that $A = \{ x < 0 : e^x \leq 0 \} = \varnothing$ and the exponential function is positive everywhere.  
A: The series expansion is
$$
e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots
$$
For $x\ge0$ we have $1$ and a bunch of nonnegative numbers, so the result is clearly positive.
For $x<0$ notice that:
$$
\frac1{e^x}=e^{-x}
$$
So positivity of $e^x$ clearly implies that $e^{-x}$ is positive.
