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?


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
$$
\frac{-x^3}{2!}\frac{1}{2-x} + \sum_{k=2}^{\infty} \frac{x^{k+1}}{(k+1)!}
= \frac{-x^3}{2!}\frac{1}{2-x} + \left( e^x - 1 - x - \frac{x^2}{2} \right)
 = e^x - \frac{2+x}{2-x} \, .
$$
