For natural $j$, let 
$$h_j:=w_j-w_{j-1},\quad s_j:=\frac{\lambda_j}{\mu_j},\quad\pi_j:=\prod_{i=1}^{j-1}s_i, 
$$
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
$$\pi_n h_n=u_n=\sum_{j=n}^\infty\frac{\pi_j}{\mu_j}\to0
$$
as $n\to\infty$ 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 $w_1<\infty$.