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had this difficult question from a textbook, and I haven't been able to figure out the solution.

say we have a sequence of bounded real numbers $a_n$ such that $2a_n \leq a_{n-1} + a_{n+1} \forall n\in\mathbb{N}$. Show that this sequence converges.

What ive tried: I did some algebra on it then tried to use the cauchy criterion, since we don't know the limit, but we know the relation between terms.

also tried moving things around and reindexing and using boundedness. but havent been able to come up with anything there either.

since bounded and monotone converges.

any help would be appreciated

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  • $\begingroup$ Getting uppercase and lowercase letters right should be rather easy. $\endgroup$ May 10, 2016 at 8:30
  • $\begingroup$ i dont know what you mean by that. $\endgroup$
    – mac5
    May 10, 2016 at 17:56
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    $\begingroup$ :-) Maybe I Underestimated The Problem. $\endgroup$ May 11, 2016 at 6:51

2 Answers 2

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Define $b_n=a_{n+1}-a_n$. Then $b_n$ is bounded and the given condition becomes $$b_{n-1}\le b_n.$$ So $b_n$ is monotonically increasing and hence it converges.

Now we have $$a_n=a_0+\sum_{i=0}^{n-1}b_i$$

Let $$\lim_{n\to\infty}b_n=c > 0$$.

Then there exists $N$ such that $b_n>c/2$ for all $n>N$, and hence $a_n$ will not be bounded for sufficiently large $n$. The case is similar for $c < 0$. So we conclude that $c=0$.

Because $b_n$ is increasing and the limit is $0$, $b_n \le 0$.

Hence $a_n$ is decreasing. Because $a_n$ is bounded, so it converges.

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We first prove by contradiction that $a_{n+1}\leq a_n$ for all $n$. So suppose that there is some $n$ such that $a_{n+1}>a_n$. Then for all $k>n$, we obtain by repeatedly using the inequality that $$\sum_{i=n+1}^{k}a_i\leq \sum_{i=n}^{k-1}a_i+\sum_{i=n+2}^{k+1}a_i.$$ From this it follows that $a_{n+1}+a_{k}\leq a_n+a_{k+1}$, hence $0<a_{n+1}-a_n\leq a_{k+1}-a_k$. By repeating this for $k=n+1,n+2,\dots,N$ (where $N$ is some integer $>k$) and adding it all up, we obtain that $$(N-n)(a_{n+1}-a_n)\leq a_{N+1}-a_{n+1},$$ hence $$a_{n+1}+(N-n)(a_{n+1}-a_n)\leq a_{N+1}.$$ But then the sequence is unbounded, which is a contradiction. Therefore, $a_{n+1}\leq a_n$ for all $n$.

Now since the sequence is decreasing and bounded, it follows that it converges.

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