Let the given series be S:
$S = \frac{1}{5}\left(\frac{1}{5} - \frac{1}{7}\right) + \left(\frac{1}{5}\right)^2\left[\left(\frac{1}{5}\right)^2 - \left(\frac{1}{7}\right)^2\right] + \left(\frac{1}{5}\right)^3\left[\left(\frac{1}{5}\right)^3 - \left(\frac{1}{7}\right)^3\right] + \dots$
This can be rewritten as:
$S = \left(\frac{1}{5}\right)^2 - \left(\frac{1}{5}\right)\left(\frac{1}{7}\right) + \left(\frac{1}{5}\right)^4 - \left(\frac{1}{5}\right)^2\left(\frac{1}{7}\right)^2 + \left(\frac{1}{5}\right)^6 - \left(\frac{1}{5}\right)^3\left(\frac{1}{7}\right)^3 + \dots$
Grouping the terms, we get:
$S = \left[\left(\frac{1}{5}\right)^2 + \left(\frac{1}{5}\right)^4 + \left(\frac{1}{5}\right)^6 + \dots\right] - \left[\left(\frac{1}{5}\right)\left(\frac{1}{7}\right) + \left(\frac{1}{5}\right)^2\left(\frac{1}{7}\right)^2 + \left(\frac{1}{5}\right)^3\left(\frac{1}{7}\right)^3 + \dots\right]$
The first bracket contains a geometric series with first term $a = \left(\frac{1}{5}\right)^2$ and common ratio $r = \left(\frac{1}{5}\right)^2$.
The second bracket also contains a geometric series with first term $a = \left(\frac{1}{5}\right)\left(\frac{1}{7}\right)$ and common ratio $r = \left(\frac{1}{5}\right)\left(\frac{1}{7}\right)$.
Using the formula for the sum of an infinite geometric series ($S = \frac{a}{1-r}$), we sum both series and find their difference to determine the value of S.
The calculated sum is: $S = \frac{5}{408}$
Thus, the sum of the infinite series is 5/408.
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