For starters,
[tex]\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k[/tex]
Consider the [tex]n[/tex]th partial sum, denoted by [tex]S_n[/tex]:
[tex]S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n[/tex]
Multiply both sides by [tex]\frac34[/tex]:
[tex]\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}[/tex]
Subtract [tex]S_n[/tex] from this:
[tex]\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)[/tex]
Solve for [tex]S_n[/tex]:
[tex]-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)[/tex]
[tex]S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)[/tex]
Now as [tex]n\to\infty[/tex], the exponential term will converge to 0, since [tex]r^n\to0[/tex] if [tex]0<r<1[/tex]. This leaves us with
[tex]\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}[/tex]
