A series representation for $e^x$ I want to show that for $x \notin \{2,3,4,\dots\}$ $$e^x = \frac{2+x}{2-x} + \sum_{k=2}^{\infty} \frac{-x^{k+1}}{k!(k-x)(k+1-x)}$$
It is pretty clear that why $x \notin \{2,3,\dots\}$. I proved that the series converges for all other $x$ by using the ratio test. My initial thoughts were to take
$$ f(x) = \frac{2+x}{2-x} + \sum_{k=2}^{\infty} \frac{-x^{k+1}}{k!(k-x)(k+1-x)}$$ and find $f'(x)$ by doing term by term by differentiation and since we know that $f(0) = 1$ we may be able to find $f(x)$. I suspect that we can not do this without proving that term by term differentiation is valid. Neverthless I tried to do that also but got stuck here.
$$ f'(x) = \frac{4}{{\left(x - 2\right)}^{2}} - \sum_{k=2}^{\infty}\left( \frac{{\left(k + 1\right)} x^{k}}{{\left(k - x + 1\right)} {\left(k - x\right)} k!} + \frac{x^{k + 1}}{{\left(k - x + 1\right)} {\left(k - x\right)}^{2} k!} + \frac{x^{k + 1}}{{\left(k - x + 1\right)}^{2} {\left(k - x\right)} k!} \right)$$
How can I show that this term by term differentiation is valid and how can I proceed further?
Can someone suggest some other ways to do it?
 A: 
Can someone suggest some other ways to do it?

Yes, one can prove the identity directly, without computing the derivative. We start with a partial fraction decomposition:
$$
\sum_{k=2}^{\infty} \frac{-x^{k+1}}{k!(k-x)(k+1-x)}
= \sum_{k=2}^{\infty}\frac{-x^{k+1}}{k!} \left( \frac{1}{k-x} - \frac{1}{k+1-x}\right) \, .
$$
The right-hand side is a convergent series of the form $\sum_{n=2}^\infty (a_n +b_n)$ with $a_n \to 0$, therefore it is allowed to re-group the terms to $a_2 + \sum_{n=2}^\infty (a_{n+1}+b_n)$. This gives
$$
\frac{-x^3}{2!}\frac{1}{2-x} + \sum_{k=2}^{\infty} \left( - \frac{x^{k+2}}{(k+1)!(k+1-x)}+\frac{x^{k+1}}{k!(k+1-x)} \right) \,.
$$
(The idea is to combine the terms with the same factor $(k+1-x)$ in the denominators.) And now the magic happens: the terms in the new sum simplify significantly, and we get
$$
\frac{-x^3}{2!}\frac{1}{2-x} + \sum_{k=2}^{\infty} \frac{x^{k+1}}{(k+1)!}
= \frac{-x^3}{2!}\frac{1}{2-x} + \left( e^x - 1 - x - \frac{x^2}{2} \right)
 = e^x - \frac{2+x}{2-x} \, .
$$
