Logarithm challenge - Find the value

$\begin{aligned} \mathscr L & = \lim_{n \to \infty} 0.16^{\log_{2.5} \left(\frac13 + \frac1{3^2} + \frac1{3^3} + \cdots + \frac1{3^n} \right)} \\ & = \lim_{n \to \infty} 0.16^{\log_{2.5} \left(\sum_{k=1}^n \left(\frac13 \right)^k\right)} \\ & = 0.16^{\log_{2.5} \left(\frac13 \cdot \frac1{1-\frac13} \right)} \\ & = 0.16^{\log_{2.5} \left(\frac12 \right)} \\ & = \left(\frac{4}{25} \right)^{-\log_\frac{5}{2} 2} \\ & = \left(\frac{2}{5} \right)^{-2 \log_\frac{5}{2} 2} \\ & = \left(\frac52 \right)^{2 \log_\frac52 2} \\ & = \left(\frac52 \right)^{\log_\frac52 4} \\ & = \boxed{4} \end{aligned}$