For natural $j$, let
$$h_j:=w_j-w_{j-1},\quad\pi_j:=\prod_{i=1}^{j-1}\frac{\lambda_i}{\mu_i},
$$
so that $\pi_1=1$, and
$$u_j:=\pi_j h_j.$$
Then the equation
$$w_n=\frac{\lambda_nw_{n+1}+\mu_nw_{n-1}+1}{\lambda_n+\mu_n}$$
for natural $n$ can be rewritten as
$$u_{n+1}=u_n-\frac{\pi_n}{\mu_n},
$$
which implies
$$u_n=u_1-\sum_{j=1}^{n-1}\frac{\pi_j}{\mu_j}. \tag{1}
$$
Also, the equality
$$w_1=\frac{1}{\mu_1}\sum_{n=0}^\infty\prod_{j=1}^n\frac{\lambda_j}{\mu_{j+1}}$$
can be rewritten as
$$u_1[=h_1=w_1]=\sum_{j=1}^\infty\frac{\pi_j}{\mu_j}.
$$
This and (1) imply
$$\pi_n h_n=u_n=\sum_{j=n}^\infty\frac{\pi_j}{\mu_j}.
$$
So, $\pi_n h_n\underset{n\to\infty}\longrightarrow0$ iff $w_1<\infty$.
Finally, the desired conclusion $$\lim_{n\to\infty}\prod_{j=1}^n\frac{\lambda_j}{\mu_j}(w_n-w_{n+1)}=0$$ can be rewritten as $\lim_{n\to\infty}\pi_n h_n=0$.
Thus, the desired conclusion holds iff $w_1<\infty$.