A line initially 1 inch long grows according to the following law, where the first term is the initial length. 
\[1+\frac{1}{4}\sqrt{2}+\frac{1}{4}+\frac{1}{16}\sqrt{2}+\frac{1}{16}+\frac{1}{64}\sqrt{2}+\frac{1}{64}+\cdots\]
If the growth process continues forever, the limit of the length of the line is: 
$\textbf{(A) } \infty\qquad \textbf{(B) } \frac{4}{3}\qquad \textbf{(C) } \frac{8}{3}\qquad \textbf{(D) } \frac{1}{3}(4+\sqrt{2})\qquad \textbf{(E) } \frac{2}{3}(4+\sqrt{2})$
We can rewrite our sum as the sum of two infinite geometric sequences.
\[1 + \frac{1}{4}\sqrt{2} + \frac{1}{4} + \frac{1}{16}\sqrt{2} + \frac{1}{16} + ... =\]
\[(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + (\frac{1}{4}\sqrt{2} + \frac{1}{16}\sqrt{2} + \frac{1}{64}\sqrt{2} + ...)\]
\[(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...)\]
We now take the sum of each of the infinite geometric sequences separately
\[(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) + \sqrt{2}(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ...) =\]
\[\frac{1}{1 - \frac{1}{4}} + \sqrt{2}(\frac{\frac{1}{4}}{1 - \frac{1}{4}})\]
\[\frac{4}{3} + \frac{\sqrt{2}}{3}\]
\[\frac{1}{3}(4 + \sqrt{2})\]
Therefore, the answer is $\fbox{(D)}$