How to show a sequence converges given that $\lim_{n \to \infty} u_n + \frac{u_{2n}}{2} = 1$ Let $u_n$ be a bounded sequence of real numbers. 
Suppose that $$\lim_{n \to \infty} u_n + \frac{u_{2n}}{2} = 1$$
Show that $u_n$ converges. 
Can someone provide some hints or insight to this problem or similar? I don't really know where to start.
 A: Let  $\epsilon>0$ be given. There exists $N$ such that $u_n-\frac{u_{2n}}2$ differs from $1$ by less than $\frac13\epsilon$ for all $n>N$.
Now if for some $n>N$, we have $|u_n-\frac23|\ge c$ with $c\ge \epsilon$, then $$\left|\frac{u_{2n}}2-\frac13\right|\ge\left|u_n-\frac23\right|-\left|\left(\frac{u_{2n}}2-\frac13\right)+\left(u_n-\frac23\right)\right|\ge c-\frac13\epsilon\ge \frac23c$$ and hence $|u_{2n}-\frac23|\ge 2c$. By induction $|u_{2^kn}-\frac23|\ge 2^kc$ for $k\in\mathbb N_0$, hence $|u_{2^kn}|\ge 2^kc-\frac23$, which contradicts boundedness of $(u_n)$. We conclude that $|u_n-\frac23|<\epsilon$ for all $n>N$. This precisely says that $\lim_{n\to\infty}u_n=\frac23$.
A: Since
$$
\lim_{n\to\infty}u_n+\frac{u_{2n}}{2}=1
$$
we have
$$
\limsup_{n\to\infty}u_n\le1-\frac12\liminf_{n\to\infty}u_n
$$
and
$$
\liminf_{n\to\infty}u_n\ge1-\frac12\limsup_{n\to\infty}u_n
$$
Subtracting, we get
$$
\limsup_{n\to\infty}u_n-\liminf_{n\to\infty}u_n\le\frac12\left(\limsup_{n\to\infty}u_n-\liminf_{n\to\infty}u_n\right)
$$
Therefore,
$$
\limsup_{n\to\infty}u_n-\liminf_{n\to\infty}u_n=0
$$
and thus, the limit exists.

To compute the limit, distribute the limit to get
$$
\lim_{n\to\infty}u_n+\frac12\lim_{n\to\infty}u_n=1
$$
and therefore,
$$
\lim_{n\to\infty}u_n=\frac23
$$
A: HINTS: Use Bolzano-Weierstrass : $u_{\phi(n)}\rightarrow l$ and by the second hypothesis $u_{2\phi(n)}\rightarrow 2(1-l):=l'$
Now can you prove that $u_n$ has only one accumulation points ?
