Is it possible to write this rational sequence : $(u_n=\frac{2^{n-1}+1}{2^n})_{n \ge 1}$

Question

using only the rational numbers : $1$; $\dfrac{1}{2}$; $\dfrac{1}{3}$; $\dfrac{1}{4}$ and the operation $+,-,\times$ ?

I get : $u_1 = 1 = \dfrac{1}{2} +\dfrac{1}{2}= \dfrac{1}{2}\left(1+1\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\right)$

$u_2 = \dfrac{3}{4} = \dfrac{1}{2} +\dfrac{1}{4}= \dfrac{1}{2}\left(1+\dfrac{1}{2}\right)= \dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}+\dfrac{1}{4}\right)$

$u_3 = \dfrac{5}{8}= \dfrac{1}{2} +\dfrac{1}{8}= \dfrac{1}{2}\left(1+\dfrac{1}{4}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)$

$u_4=\dfrac{9}{16}=\dfrac{1}{2} +\dfrac{1}{16}=\dfrac{1}{2}\left(1+\dfrac{1}{8}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{8}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\times \dfrac{1}{4}\right)$

etc...

$u_n = \dfrac{2^{n-1}+1}{2^n}= \dfrac{1}{2} +\dfrac{1}{2^{n}}=\dfrac{1}{2}\left(1+\dfrac{1}{2^{n-1}}\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2^{n-1}}\right)$ and following the value of $n$ we can make appear several products of $\dfrac{1}{2}$ and $\dfrac{1}{4}$.

Can we find a better form for $u_n$ ?

Moreover it seems possible to write the terms of this sequence (for $n>1$) using only once the rational numbers $1$; $\dfrac{1}{2}$; $\dfrac{1}{3}$; $\dfrac{1}{2^n}$ :

$u_2 = \dfrac{1}{3}\left(1+\dfrac{1}{2}\right)+\dfrac{1}{4}$

$u_3 = \dfrac{1}{3}\left(1+\dfrac{1}{2}\right)+\dfrac{1}{8}$

etc...

$u_n = \dfrac{2^{n-1}+1}{2^n}= \dfrac{1}{2} +\dfrac{1}{2^{n}}=\dfrac{1}{3}\left(1+\dfrac{1}{2}\right)+\dfrac{1}{2^{n}}$

Thanks in advance !

Answer 1

As mentioned in the comments, you can use a recurrence, which needs only $\frac{1}{2},+$ and $\times$:
Since $u_n=\frac{1}{2}+\frac{1}{2^n} \Rightarrow \frac{1}{2^n}=u_n-\frac{1}{2}$, we have $u_{n+1}=\frac{1}{2}+\frac{1}{2^{n+1}}=\frac{1}{2}(1+\frac{1}{2^n})$ and the recurrence relation $u_{n+1}=\frac{1}{2}(\frac{1}{2}+u_n)$. By $u_1=\frac{1}{2}+\frac{1}{2}$ we are finished.
The first n are:
$u_1=\frac{1}{2}+\frac{1}{2}$,
$u_2=\frac{1}{2}(\frac{1}{2}+u_1)=\frac{1}{2}(\frac{1}{2}+\frac{1}{2}+\frac{1}{2})$,
$u_3=\frac{1}{2}(\frac{1}{2}+u_2)=\frac{1}{2}(\frac{1}{2}+\frac{1}{2}(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}))$.