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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.

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3 Answers 3

up vote 2 down vote accepted

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, $$ \lim_{n\to\infty}u_n=\frac23 $$

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this seem really cool but could you elaborate further on how to get the first two relations? From there on I understand everything :-) –  Ant Apr 24 at 16:38
    
@Ant: Show that $$\limsup_{n\to\infty}(-u_n)=-\liminf_{n\to\infty}u_n$$ which incidentally shows that $\displaystyle\liminf_{n\to\infty}(-u_n)=-\limsup_{n\to\infty}u_n$. Show that $$\limsup_{n\to\infty}(u_{2n})\le\limsup_{n\to\infty}u_n$$ which, using the previous equalities, also shows that $\displaystyle\liminf_{n\to\infty}(u_{2n})\ge\liminf_{n\to\infty}u_n$. Apply these to $$u_n=1-\frac12u_{2n}$$ –  robjohn Apr 24 at 16:54
    
thank you. very elegant! :) –  Ant May 4 at 9:11

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 $\lim_{n\to\infty}u_n=\frac23$.

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(+1) I think this is in essence the same as what I did, though I had a longer answer that I overwrote that implies that $$\frac23=\frac12\frac54\frac{17}{16}\frac{257}{256}\cdots$$ –  robjohn Apr 23 at 21:36
    
This also seem very cool though I have two problems: 1) At the very beginning, should be $u_n + \frac{u_{2n}}{2}$ instead? 2) I can't prove why is that the first inequality holds. why $$\left|\frac{u_{2n}}2-\frac13\right|\ge\left|u_n-\frac23\right|-\left|\left( \frac{u_{2n}}{2}-\frac13\right)+\left(u_n-\frac23\right)\right|$$ ? From there on I think I got it though! –  Ant Apr 24 at 16:45

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 ?

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No I couldn't. I can show $u_{4\phi(n)} \to 2(2l-1) := l''$ and $l'' = l' = l$ if and only if $l = 2/3$. We can probably continue along these lines with $u_{8\phi(n)}$ but I don't see how to conclude... –  Ant Apr 23 at 20:05
    
@Ant You are on the right road, $u_{2\phi(n)}+\frac{u_{4\phi(n)}}{2}\rightarrow l$ then $l'+\frac{l'}{2}=1$. So $l'=\frac{2}{3}$ and thus $l=\frac{2}{3}$. You can conclude that $u_n$ has only one accumulation points and therefore converge to $l=\frac{2}{3}$. –  user142836 Apr 23 at 21:29
    
I'm sorry but I can't follow you. If you could expand your answer and add more details (and completing the proof) it would be greatly appreciated! :-) –  Ant Apr 24 at 16:48

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