A recursive formula to approximate $e$. Prove or disprove.

It can be proved by induction that $x_n = \frac{1}{(n-1)!}$

$x_1$ and $x_2$ satisfy this, and by doing induction,

$$x_{m+1} = x_{m}-\frac{1}{m}x_{m-1} = \frac{1}{(m-1)!} - \frac{1}{m}\frac{1}{(m-2)!} = \frac{1}{m!} $$

Thus you series reduces to the ordinary series $\sum_{n=0}^{\infty}\frac{1}{n!}$ for $e$

The recursion is satisfied by $$x_n=\frac{1}{\Gamma (n)}$$ So, $$\sum_{i=1}^p x_i=\frac{e\, p \,\Gamma (p,1)}{\Gamma (p+1)}$$ where appears the incomplete gamma function and the limit is effectively $e$.

The first terms of the sequence are $$1,1,\frac12,\frac16,\frac1{24},\frac1{120}\cdots$$which you recognize without surprise.

Indeed, with $x_n=\dfrac1{(n-1)!}$ you have

$$x_n=nx_{n+1},$$ $$x_{n-1}=n(n-1)x_{n+1},$$ hence $$x_n-\frac nn(n-1)x_{n-1}=x_{n+1}.$$

Nothing new, sorry.