According to Taylor Series, $$f(x) = \sum_{n=0}^\infty \dfrac{f^{(n)}(a)}{n!}*(x-a)^n $$
However, $\dfrac{1}{x}$,$\dfrac{1}{x^2}$, etc. are not applicable. I tried to do the following:
$$ \dfrac{1}{x}=\sum_{n=0}^\infty (1-x)^n $$
Which I assume $a$ is $1$. Now, if $x$ is a positive integer, then a problem exists! The final value of the equation above must also be an integer, yet the answer is impossible to be an integer. Otherwise it is basically true. How come it fails if $x$ is integer?
Of course, the Taylor Series will not work when the function is an error function, such as $\ln x$ (where $x$ is negative).
In what situation will Taylor Series be wrong other than error functions?
Best Answer
For a Taylor Series of a function:
$f(x) = \displaystyle\sum\limits_{k=0}^{+\infty}c_k (x-a)^k = c_0 + c_1(x-a) + c_2(x-a)^2 + \dots$
The equality holds for $x-a$ in the radius of convergence:
$\displaystyle\frac{1}{R} = \limsup\limits_{k \rightarrow \infty} \left|c_k\right|^{\large{\frac{1}{k}}}$ (Cauchy-Hadamard formula)
where $R$ is the radius of convergence of the given function, or more concretely, the power series converges to the function for all $x$ that satisfies: $\displaystyle\left|\,x-a\, \right| < R$, where $R\in [0,+\infty]$. $\hspace{1cm}\blacksquare$
In your example, for $\displaystyle f(x) = \frac{1}{x}$, we have:
$\displaystyle f(x) = \sum\limits_{k=0}^{+\infty} \frac{(-1)^k}{a^k}(x-a)^k$
so $\displaystyle c_k = \frac{(-1)^k}{a^k}$.
Now for radius of convergence:
$\displaystyle\frac{1}{R} = \limsup\limits_{k\rightarrow\infty} \frac{1}{\sqrt[k]{|a|^k}} = \frac{1}{|a|} \implies R = |a|.$
So our function is convergent for: $|x-a| < |a| \Longleftrightarrow \left\{ \begin{array}{l l} 0 < x < 2a & \quad \text{for $a>0$}\\ 2a < x < 0 & \quad \text{for $a<0$} \end{array} \right.$
Now, you wanted $a = 1$, so our series is convergent for $\displaystyle\boxed{0 < x < 2}$. For all other $x$, the series is divergent.