How can I define $e^x$ as the value of infinite series? I understand the definition of $e^x$ by limit. But I do not know how to come up with:
$$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$$
without using Taylor series. more explicitly without using calculus. how do we know if a function can be expressed as series or not ?
 A: One definition of the exponential function is the limit
$$
\lim_{n\to\infty} \Big(1 + \frac{x}{n}\Big)^n=e^x.
$$
Let $P_n(x)$ denote the polynomial $(1+x/n)^n$, so that $e^x=\lim_{n\to\infty}  P_n(x)$; I will show that
$$
\lim_{n\to\infty} P_n(x) = \sum_{n=0}^\infty \frac{x^n}{n!}.
$$
If you expand out $P_n(x)$ using the binomial theorem, you'll find that
$$
P_n(x) = \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k}.
$$
Now
\begin{align}
\frac{1}{n^k}\binom{n}{k} 
&= \frac{1}{n^k}\cdot\frac{n}{k}\cdot\frac{n-1}{k-1}\cdots\frac{n-k+1}{1} \\
&= \frac{1}{k}\cdot\frac{1-1/n}{k-1}\cdots\frac{1-(k-1)/n}{1}.
\end{align}
Since $k$ is fixed, $\lim_{n\to\infty}\binom{n}{k}/n^k=\frac{1}{k!}$. It follows that
$$
\lim_{n\to\infty} P_n(x) = \sum_{k=0}^\infty\lim_{n\to\infty}\binom{n}{k}\frac{x^k}{n^k}
= \sum_{k=0}^\infty\frac{x^k}{k!}.
$$
You should look at the Wikipedia article about characterizations of the exponential function; it has five.
As for your question about whether a function can be expressed as a series or not, to answer it I believe you need to say something about calculus. What I mean is that if a "nice" function $f(x)$ has a series representation at a point $a$ then the series is given by
$$
f(x)=\sum_{k=0}\frac{f^{(k)}(a)}{k!}(x-a)^k,
$$
where $f^{(k)}(a)$ is the $k$th derivative of $f$ at $a$.
A: If you take the fact that $(e^x)'=e^x$ then by taylor series, we can expand around $0$, taking into account that each $n$-th derivative at $0$ is $1$.
Then $e^x=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{1}{n!}x^n=\sum_{n=0}^\infty\frac{x^n}{n!}$
Lets try this without Taylor series. A function that can be expressed by a real power series is called real analytic. All that is needed is that all derivatives are greater than or equal to $0$. Clearly this holds for $e^x$.
So let us write $e^x=\sum_{n=0}^\infty a_nx^n$ taking derivatives we get:
$\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty a_nx^n$
Comparing coefficients of each power of $x$ we get:
$a_{n+1}=\frac{1}{n+1}a_n$ and we know $a_0=1$ (because $e^0=1$) this gives as $a_n=\frac{1}{n!}$
Thus $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
