The value of $$0.16^{\log_{2.5}\left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \infty\right)}$$ is __________

We have to evaluate the expression

$$0.16^{\log_{2.5}\left(\dfrac{1}{3} + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \ldots \infty\right)}.$$

First, focus on the infinite series that appears inside the logarithm. The series

$$\dfrac{1}{3} + \dfrac{1}{3^2} + \dfrac{1}{3^3} + \ldots$$

is an infinite geometric progression. For an infinite GP with first term $$a$$ and common ratio $$r$$ (where $$|r| < 1$$), the sum is given by the well-known formula

$$S = \dfrac{a}{1 - r}.$$

Here the first term is $$a = \dfrac{1}{3}$$ and the common ratio is $$r = \dfrac{1}{3}.$$ Substituting into the formula, we obtain

$$S \;=\; \dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} \;=\; \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} \;=\; \dfrac{1}{3}\times\dfrac{3}{2} \;=\; \dfrac{1}{2}.$$

So the entire geometric series sums to $$\dfrac{1}{2}.$$ Hence the exponent becomes

$$\log_{2.5}\!\left(\dfrac{1}{2}\right).$$

Let us denote this exponent by a single symbol for clarity. Set

$$x = \log_{2.5}\!\left(\dfrac{1}{2}\right).$$

By definition of the logarithm, this means

$$(2.5)^{x} = \dfrac{1}{2}.$$

Next, examine the base of the main power, $$0.16.$$ Notice that

$$0.16 = \dfrac{16}{100} = \dfrac{4}{25} = \left(\dfrac{2}{5}\right)^{2}.$$

Observe also that $$2.5 = \dfrac{5}{2}.$$ Hence, because $$\dfrac{2}{5}$$ is the reciprocal of $$\dfrac{5}{2},$$ we can rewrite the base in terms of $$\dfrac{5}{2}$$:

$$0.16 = \left(\dfrac{2}{5}\right)^{2} = \left(\dfrac{5}{2}\right)^{-2}.$$

Therefore, the original expression becomes

$$\bigl(0.16\bigr)^{x} \;=\; \left[\left(\dfrac{5}{2}\right)^{-2}\right]^{x} \;=\; \left(\dfrac{5}{2}\right)^{-2x}.$$

But from the definition of $$x$$ we already have

$$(\tfrac{5}{2})^{x} = \dfrac{1}{2}.$$

Using the law of exponents, specifically $$\bigl(a^{m}\bigr)^{n} = a^{mn},$$ we can rewrite

$$\left(\dfrac{5}{2}\right)^{-2x} \;=\; \Bigl[\bigl(\tfrac{5}{2}\bigr)^{x}\Bigr]^{-2} \;=\; \left(\dfrac{1}{2}\right)^{-2}.$$

Now employ the rule $$a^{-n} = \dfrac{1}{a^{\,n}},$$ or equivalently $$\left(\dfrac{1}{k}\right)^{-n} = k^{n}.$$ Here $$k = 2$$ and $$n = 2,$$ so

$$\left(\dfrac{1}{2}\right)^{-2} = 2^{2} = 4.$$

Hence, the value of the original expression is

$$4.$$

So, the answer is $$4$$.