Clarification on Discrete-time convolution and signal shifting What are $y_1[n]=x_1[n]*h[n]$ and $y_2[n]=x_2[n]*h[n]$?
  $$x_1[n]=(0.1)^nu[n],\quad x_2[n]=(0.2)^n,\quad h[n]=(0.3)^nu[n]$$


I am confused about how to calculate $y_{1}[n]$. The formula for discrete convolution is:
 $$\sum_{m=-\infty}^{\infty}x[m]h[n-m] = \sum_{m=-\infty}^{\infty}h[m]x[n-m]$$.

My attempt:
I chose to shift $x_{1}[n]$ along m-axis:
$$y_{1}[n] = \sum_{m=0}^{n} (0.3)^{m}(0.1)^{n-m}$$
$$y_{1}[n] =(0.1)^{n}\sum_{m=0}^{n}(\frac{0.3}{0.1})^{m}$$
$$y_{1}[n] = (0.1)^{n}\sum_{m=0}^{n}(3)^{m}$$

This gives me $y_{1}[n] = \infty$. However, when I shift $h[n]$ along the m-axis I get $ y_{1}[n]$ as a finite value. My understanding of discrete time convolution was that shifting either $x[n]$ or $h[n]$ would result in the same answer.
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 $n&lt;0$.