Convergence of $ z_n=\frac{1}{n} \left[ x_1\left(1+\frac{x}{n}\right)^n + x_2\left(1+\frac{x}{n-1}\right)^{n-1} + \dots + x_n(1+x) \right] $

Question

Question

If there is some convergent sequence $\{x_n\} \to x$ where each $x_n>0$ then $$ z_n=\frac{1}{n} \left[ x_1\left(1+\frac{x}{n}\right)^n + x_2\left(1+\frac{x}{n-1}\right)^{n-1} + \dots + x_n(1+x) \right] $$ What is $\lim_{n \to \infty} z_n=z$

My attempt

Initially I thought we could use the converse of Cauchy's first limit theorem (i.e.if a positive sequence $\{a_n\} \to l$ then $\lim_{n \to \infty}\sum_{i=1}^n \frac{a_i}{n}=l$). But that gives $z= \lim_{n \to \infty} x_n(1+x)=x(1+x)$. The answer that is listed is given as $z=xe^x$ so this seems to be wrong.

Also, I'm not sure if the converse of that theorem is even valid. So I think this approach to the question must be wrong.

Alternatively. I assume we need to convert the following summation to a Riemann integral but I'm unsure how to proceed.

Let $\lim_{n \to \infty} z_n=z$ $$z=\lim_{n \to \infty}\sum_{i=1}^n \frac{1}{n} x_i\left(1+\frac{i}{n-i}\right)^{n-i}$$

I know require some function in terms of $i/n$ but even trying some change of variables I do not know how to deal with the exponent term.

Any helps or hints on how to proceed would be appreciated.