I'm trying to prove that the following recursive sequence converges to 1:$$x_1=0$$$$x_{n+1}=x_n-2\frac{e^{x_n}-ex_n}{e^{x_n}-e}$$The sequence has no fixed points, but $\lim_{x\to1}\frac{e^x-ex}{e^x-e}=0$ (and this is the only value of $x$ for which this is true). I'm pretty sure that means that if the sequence does converge, it must converge to $1$, but "I'm pretty sure" isn't much of a proof. I don't see any way to find an explicit form for the sequence (and I suspect there isn't one). The sequence isn't monotonic, since it oscillates above and below $1$. I can't find any way to satisfy Cauchy's criterion for convergence; the difference between two arbitrary terms $x_m,x_n$ ($n>m$) is $-2\sum_{i=m}^{n-1}\frac{e^{x_i}-ex_i}{e^{x_i}-e}$, and I don't see any way to show that this sum is arbitrarily small unless I already know that each $x_i$ is close to $1$. I'm not sure what else I can do to prove that the sequence converges, either in general or specifically to $1$.
For some context, I'm working with a modified Newton's Method, that uses the sequence$$x_{n+1}=x_n-p\frac{f(x_n)}{f'(x_n)},$$where $p$ is the multiplicity of the root we're trying to find. $f(x)=e^x-ex$, which has a single root at $x=1$ with multiplicity $2$. I'm being asked to show that "the convergence of the sequence is quadratic", and I can do that if I assume the sequence converges to the root in the first place, but I don't think that's a reasonable assumption to make, and I wasn't explicitly told to assume it. We did prove that the ordinary Newton's Method converges, but only in the case where $f'$ is nonzero in a neighbourhood around the root, and I don't see any way to extend that proof to apply here. It's been years since I took a course that dealt with convergence of sequences in detail, and I don't really remember what approach to take to solve a problem like this. I would appreciate any assistance!