Clarification on Discrete-time convolution and signal shifting
You're right that both results should be identical. The mistake lies in your conclusion that $y_1[n]=\infty$ in the first case, that's not true. You get

$$y_1[n]=(0.1)^nu[n]\sum_{m=0}^n3^m=(0.1)^nu[n]\frac{3^{n+1}-1}{3-1}=\ldots\tag{1}$$

Note that you must also include the unit step $u[n]$ in $(1)$ because the response is zero for $n&lt;0$.