\( T_{1}=8, T_{2}=\frac{8}{\sqrt{2}} \) 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 converges to \( 16 + 8 \sqrt{2} \), we need to recognize the terms provided for the series. The first term \( T_1 = 8 \) can be considered as the initial term of a geometric series. The common ratio \( r \) can be found using the next term: \[ r = \frac{T_2}{T_1} = \frac{\frac{8}{\sqrt{2}}}{8} = \frac{1}{\sqrt{2}} \] This means we can set up our series as: \[ S = T_1 + T_2 + T_3 + \ldots = 8 + 8 \left(\frac{1}{\sqrt{2}}\right) + 8 \left(\frac{1}{\sqrt{2}}\right)^2 + \ldots \] Since this is an infinite geometric series with first term \( a = 8 \) and common ratio \( r = \frac{1}{\sqrt{2}} \), we can calculate the sum using the formula for the sum of an infinite geometric series \( S = \frac{a}{1 - r} \), provided \( |r| < 1 \): \[ S = \frac{8}{1 - \frac{1}{\sqrt{2}}} \] To simplify, we calculate the denominator: \[ 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \] Thus, the sum \( S \) becomes: \[ S = \frac{8 \times \sqrt{2}}{\sqrt{2} - 1} \] To make this easier, multiply the numerator and denominator by \( \sqrt{2} + 1 \): \[ S = \frac{8 \sqrt{2} (\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{8 \sqrt{2} (\sqrt{2} + 1)}{2 - 1} = 8 \sqrt{2} (\sqrt{2} + 1) = 8 \cdot 2 + 8 \sqrt{2} = 16 + 8 \sqrt{2} \] Thus, we have proven that the sum of the series to infinity is \( 16 + 8 \sqrt{2} \).