\( T_{1}=8, T 12=\frac{8}{\sqrt{2}} \) Prove, without the use of a calculator, that the sum of the series to infinity is \( 16+8 \sqrt{2} \)

To prove that the sum of the series to infinity is \( 16 + 8\sqrt{2} \), we start by identifying the series. Given \( T_1 = 8 \) and \( T_{12} = \frac{8}{\sqrt{2}} \), we're dealing with a geometric series where each term is derived from the previous term by a common ratio. The common ratio \( r \) can be calculated as: \[ r = \frac{T_{12}}{T_{1}} = \frac{\frac{8}{\sqrt{2}}}{8} = \frac{1}{\sqrt{2}}. \] The first term \( a = T_1 = 8 \). For a geometric series, the sum to infinity \( S \) can be calculated using the formula: \[ S = \frac{a}{1 - r} \] as long as \( |r| < 1 \). Substituting our values, we get: \[ S = \frac{8}{1 - \frac{1}{\sqrt{2}}} = \frac{8\sqrt{2}}{\sqrt{2} - 1}. \] Next, we rationalize the denominator: \[ S = \frac{8\sqrt{2}(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{8\sqrt{2}(\sqrt{2} + 1)}{2 - 1} = 8\sqrt{2}(\sqrt{2} + 1) = 16 + 8\sqrt{2}. \] Thus, we have proven that the sum of the series to infinity is indeed \( 16 + 8\sqrt{2} \).