You can factor the 32 out of the sum:
[tex]\displaystyle \sum_{m=1}^\infty 32 \cdot \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1}[/tex]
We can also change the index as follows
[tex]\displaystyle 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m}[/tex]
Now, we have a theorem that states that the series
[tex]\displaystyle \sum_{m=1}^\infty a^m[/tex]
converges if and only if [tex]|a|<1[/tex], and in this case we have
[tex]\displaystyle \sum_{m=1}^\infty a^m = \dfrac{1}{1-a}[/tex]
This is your case, because you have
[tex]|a|=\dfrac{1}{2}<1[/tex]
which implies that your series converges, and the value is
[tex]\displaystyle 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m} = 32 \cdot \dfrac{1}{1-\frac{1}{2}} = 32\cdot 2 = 64[/tex]
