Logarithm challenge - Find the value

Algebra Level 3

lim⁡n→∞0.16log⁡2.5(13+132+133+⋯+13n)= ?\large \lim_{n \to \infty} 0.16^{\log_{2.5}\left( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots + \frac{1}{3^n} \right)} = \, ?

Nithin Konidala by Nithin Konidala
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