\( \sum _ { n = 1 } ^ { \infty } \frac { ( x - 5 ) ^ { n } } { n ^ { 2 } \cdot 7 ^ { n } } = \)

To find the sum of the series \( \sum _ { n = 1 } ^ { \infty } \frac { ( x - 5 ) ^ { n } } { n ^ { 2 } \cdot 7 ^ { n } } \), we can recognize that this is a power series in the form of \( \sum_{n=1}^{\infty} a_n \), where \( a_n = \frac{(x-5)^n}{n^2 \cdot 7^n} \). This series can be manipulated into a more standard form. We can rewrite it as: \[ \sum_{n=1}^{\infty} \frac{(x-5)^n}{n^2} \cdot \frac{1}{7^n} = \sum_{n=1}^{\infty} \frac{(x-5)^n}{n^2} \cdot \left(\frac{1}{7}\right)^n \] The term \( \frac{(x-5)}{7} \) can be denoted as \( y \), so we have: \[ \sum_{n=1}^{\infty} \frac{y^n}{n^2} \] This series resembles the Taylor series for the function \( -\ln(1-y) \), where \( |y| < 1 \), but it actually converges to \( \text{Li}_2(y) \) or the dilogarithm function, which is defined for this form. Thus, the original series summation can be expressed as: \[ \sum_{n=1}^{\infty} \frac{(x-5)^n}{n^2 \cdot 7^n} = \text{Li}_2\left(\frac{x-5}{7}\right) \] This series converges for \( \left|\frac{x-5}{7}\right| < 1 \), or equivalently, \( 5 - 7 < x < 5 + 7 \), which simplifies to \( -2 < x < 12 \).