This series converges.
[tex]S=\dfrac14+\dfrac3{16}+\dfrac1{64}+\dfrac3{256}+\dfrac1{1024}+\cdots[/tex]
[tex]S=\left(\dfrac34+\dfrac3{16}+\dfrac3{64}+\dfrac3{256}+\dfrac3{1024}+\cdots\right)-\left(\dfrac12+\dfrac1{32}+\dfrac1{512}+\cdots\right)[/tex]
[tex]S=\displaystyle\sum_{n=1}^\infty\frac3{4^n}-\sum_{n=0}^\infty\frac1{2^{4n+1}}[/tex]
Both component series are geometric with ratios less than 1, so they both converge.
[tex]\displaystyle\sum_{n=1}^\infty\frac3{4^n}=1[/tex]
[tex]\displaystyle\sum_{n=0}^\infty\frac1{2^{4n+1}}=\frac12\sum_{n=0}^\infty\frac1{16^n}=\frac8{15}[/tex]
So we have
[tex]S=\dfrac7{15}[/tex]
