9514 1404 393
Answer:
2n/2^n
Step-by-step explanation:
It appears that the terms are ...
[tex]\dfrac{1}{1},\ \dfrac{2}{2},\ \dfrac{3}{4},\ \dfrac{4}{8},\ \dots[/tex]
That is, the numerator is counting numbers, and the denominator is powers of 2.
[tex]f[n]=\dfrac{n}{2^{(n-1)}}\quad\text{or}\quad f[n]=2n\cdot2^{-n}[/tex]
