We know that the exponential function $e^x$ and the natural logarithm $log(x)$ are inverses of each other, thus $e^{log(x)}=x$ and $log(e^x) = x$. We can use Taylor Series to
determine a series for each of these functions.
Taylor Series of exponential function
We know the following:
$$
\begin{aligned}
f^{(1)}(x) & = {d \over dx}\left( e^x \right) = e^x\\
f^{(n)}(x) & = {d^{(n)} \over dx}\left( e^x \right) = e^x\\
e^0 & = 1\\
f^{(n)}(0) & = {d^{(n)} \over dx}\left( e^0 \right) = e^0 = 1\\
\end{aligned}
$$
Therefore the Taylor Series of the exponent function is represented by:
$$
\begin{aligned}
e^x & = \sum_{j=0}^\infty f^{(j)}(0) \cdot {{x^j} \over j!}\\
& = \sum_{j=0}^\infty 1 \cdot {{x^j} \over j!}\\
e^x & = \sum_{j=0}^\infty {{x^j} \over j!}\\
\end{aligned}
$$
If we apply this to $x=1$ then we get the following:
$$
\begin{aligned}
e^1 & = \sum_{j=0}^\infty {{1^j} \over j!}\\
e^1 & = \sum_{j=0}^\infty {1 \over j!}\\
e & = 1 + {1 \over 1!} + {1 \over 2!} + {1 \over 3!} + {1 \over 4!} + ... + {1 \over n!} + .....\\
& = 1 + 1 + {1 \over 2} + {1 \over 6} + {1 \over 24} + {1 \over 120} + ......\\
& = 2.718281828459045...\\
\end{aligned}
$$
We can in fact use this to determine $e$ to any number of digits. The first 100 digits determined by this method are:
$$
\begin{aligned}
e = 2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742....\\
\end{aligned}
$$
Taylor Series for natural logarithm
When we want to work the natural logarithm we have a problem if we try to determine $log(0)$ seeing that it is undefined.
We can overcome this by setting $a=1$ in the following:
$$
\begin{aligned}
f(x) & = \sum_{j=0}^\infty f^{(j)}(a) \cdot {{(x-a)^j} \over j!}\\
f(x) & = \sum_{j=0}^\infty f^{(j)}(1) \cdot {{(x-1)^j} \over j!}\\
log(1) & = 0\\
log^{(1)}(a) & = {1 \over a}\\
log^{(1)}(1) & = {1 \over 1} = 1\\
log^{(2)}(a) & = -{1 \over {a^2}}\\
log^{(2)}(1) & = -{1 \over {1^2}} = -1\\
log^{(3)}(a) & = {2 \over {a^3}}\\
log^{(3)}(1) & = {2 \over {1^3}} = 2\\
log^{(n)}(a) & = (-1)^{n+1} \cdot {{(n-1)!} \over {a^n}}\\
log^{(n)}(1) & = (-1)^{n+1} \cdot {{(n-1)!} \over {1^n}} = (-1)^{n+1} \cdot (n-1)!\\
\therefore\, log(x) & = \sum_{j=1}^\infty (-1)^{j+1} \cdot (j-1)! \cdot {{(x-1)^j} \over j!}\\
& = \sum_{j=1}^\infty (-1)^{j+1} \cdot {{(x-1)^j} \over j}\\
\therefore\, log(x+1) & = \sum_{j=1}^\infty (-1)^{j+1} \cdot {{x^j} \over j}\\
& = x - {{x^2} \over 2} + {{x^3} \over 3} - {{x^4} \over 4} + ..... (-1)^{n+1} {{x^n} \over n} + .....
\end{aligned}
$$
We can however see a problem will arise if $x > 1$. Let's take $x = 1+y$, then $x^n = {(1+y)}^n = 1 + n \cdot y + ....$. If we let $n$ goes towards
positive infinity $1 + n \cdot y$ will go toward infinity and seeing that the difference between $x^n$ and $x^{n-1}$ does not converge the sum does not as well.
We can see if we make $x=0$ then
$$
\begin{aligned}
log(1+0) & = 0 - {{0^2} \over 2} + {{0^3} \over 3} - {{0^4} \over 4} + ..... (-1)^{n+1} {{0^n} \over n} + .....\\
log(1) & = 0\\
\end{aligned}
$$