Show that $nu_n$ converges to $1$. $\Bbb{K}=\Bbb{R}$ or $\Bbb{C}$

Let $(u_n)\in\Bbb{K}^{\Bbb{N}}$ be a sequence such that $(n(u_n+u_{2n}))_{n\in \Bbb{N}}$ converges to $\frac{3}{2}$ and $u_n\rightarrow 0$.
It is asking to prove that  $(nu_n)$ converges  to $1$.

If I denote $v_n=nu_n$, we have $v_n+\frac{v_{2n}}{2}\rightarrow 3/2$.
If $V_n$ is positive then I can conclude that $v_n$ is bounded and use some result I have in my course.
Any Ideas How can I continue?
 A: Put $v_n=nu_n$, and $\displaystyle v_n+\frac{v_{2n}}{2}=3/2+a_n$, with $a_n\to 0$ if $n\to +\infty$.
We have hence $\displaystyle v_n=-\frac{v_{2n}}{2}+3/2+a_n$. Replace $n$ by $2n$, and use the expression for $v_{2n}$,  we get
$$v_n=(-\frac{1}{2})^2v_{2^2n}+\frac{3}{2}(1-\frac{1}{2})+a_n+(-\frac{1}{2})a_{2n}$$
By induction on $k$, we have for all $n$:
$$v_n=(-\frac{1}{2})^k v_{2^k n}+\frac{3}{2}(1-\frac{1}{2}+\cdots+(-\frac{1}{2})^{k-1}))+a_n+(-\frac{1}{2})a_{2n}+\cdots+(-\frac{1}{2})^{k-1})a_{2^{k-1}n}$$
Now  fix $n$, and let $k\to +\infty$, ; as $u_m\to 0$ as $m\to+\infty$, we get that $\displaystyle (-\frac{1}{2})^k v_{2^k n}\to 0$, and
$$v_n=1+\sum_{j\geq 0}(-1)^{j}\frac{a_{2^jn}}{2^j}$$
As $a_n\to 0$ if $n\to +\infty$, it is now easy to see that $\displaystyle \sum_{j\geq 0}(-1)^{j}\frac{a_{2^jn}}{2^j} \to 0$ as $n\to \infty$, and we are done. 
A: Let be $v=\lim v_n$ (of course when $n$ go to $+\infty$). Now you since $v_{2n}$ is a subsecuence of $v_n$ you can conclude $v=\lim v_n=\lim v_{2n}$
You now $\lim v_n+\frac{v_{2n}}{2}=\frac{3}{2}$, but you also know:
$$\lim v_n+\frac{v_{2n}}{2}=\lim v_n+\lim \frac{v_{2n}}{2}=\lim v_n+ \frac{\lim v_{2n}}{2}=v+\frac{v}{2}=\frac{3}{2}v$$
Then
$$\frac{3}{2}v=\frac{3}{2}$$
And then
$$v=1$$
And you conclude $\lim v_n=\lim nu_n=v=1$.
