Arithmetic-Geometric Series

note that sums to infinity is $\frac{a}{1-r}$ , where a is the first number of the series and that r is the ratio (exponent).

$S=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}...$ $S=\displaystyle \sum_{k=1}^\infty (\frac{1}{2})^k+\displaystyle \sum_{k=1}^\infty (\frac{1}{2})^{k+1}+\displaystyle \sum_{k=1}^\infty (\frac{1}{2})^{k+2}...$ $S=1+\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...$ $S=\displaystyle \sum_{k=1}^\infty (1/2)^{k-1}$ $S=2$

Problem needs more terms. Could be interpreted as $\sum_{n=0}^\infty\frac{5^n}{2^{4n+1}}+\sum_{n=0}^\infty\frac{2\cdot5^n}{2^{4n+2}}+\sum_{n=0}^\infty\frac{3\cdot5^n}{2^{4n+3}}+\sum_{n=0}^\infty\frac{4\cdot5^n}{2^{4n+4}}=\frac{26}{11}$