Sum of Infinite Series

The $n$-th term is $T_n = \left(\sum_{k=0}^{n-1} \frac{1}{3^k}\right) \frac{1}{4^{n-1}}$.

Using GP sum formula: $\sum_{k=0}^{n-1} \frac{1}{3^k} = \frac{1 - (1/3)^n}{1 - 1/3} = \frac{3}{2}\left(1 - \frac{1}{3^n}\right)$ $T_n = \frac{3}{2}\left(1 - \frac{1}{3^n}\right)\frac{1}{4^{n-1}} = \frac{3}{2}\left(\frac{1}{4^{n-1}} - \frac{1}{3 \cdot 12^{n-1}}\right)$

Sum of infinite series: $S = \sum_{n=1}^\infty T_n = \frac{3}{2} \left( \sum_{n=1}^\infty \frac{1}{4^{n-1}} - \frac{1}{3} \sum_{n=1}^\infty \frac{1}{12^{n-1}} \right)$ $S = \frac{3}{2} \left( \frac{1}{1 - 1/4} - \frac{1}{3} \cdot \frac{1}{1 - 1/12} \right) = \frac{3}{2} \left( \frac{4}{3} - \frac{4}{11} \right)$ $S = \frac{3}{2} \cdot 4 \cdot \left(\frac{1}{3} - \frac{1}{11}\right) = 6 \times \frac{8}{33} = \frac{16}{11}$