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Let $ a_n $ be a bounded sequence of real numbers such that $ 2 a_n \leq a_{n-1} + a_{n+1} $ for $ n = 2,3,...$ . Show that $ a_{n+1} - a_{n} \to 0 $

mr.stealyourgirl
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1 Answers1

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Rewrite the inequality as $a_{n+1}-a_n \ge a_n-a_{n-1}$. Let $b_{n}=a_{n+1}-a_n$.

Then $(b_n)$ is an increasing sequence which is bounded above, so has a limit.

If that limit is not $0$, then the sequence $(a_n)$ is unbounded. Suppose for example that $(b_n)$ has limit $l\gt 0$. Then after a while $b_n\gt \frac{l}{2}$, which means that the sequence $(a_n)$ is not bounded above.

Similar but somewhat trickier reasoning deals with $l\lt 0$. There is an absolute bound $B$ on the elements of $(a_n)$. Choose a very large $n$, and go downwards.

André Nicolas
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  • I can't understand why $ b_n \to l > 0 $ implies that $a_n$ is not bounded – mr.stealyourgirl Sep 02 '13 at 20:21
  • @Dman-pwns The differences between the $a_n$ are all positive (after a certain point) and greater than some number, so the $a_n$ shoot off to infinity. – Potato Sep 02 '13 at 20:23
  • There is an $N$ such that if $n\ge N$ then $a_{n+1}-a_n\gt \frac{l}{2}$. So $a_{N+1}\gt a_N+\frac{l}{2}$. So $a_{N+2}\gt a_N+l$. So $a_{N+3}\gt a_N+\frac{3l}{2}$. So $a_{N+4}\gt a_N+2l$. And so on. – André Nicolas Sep 02 '13 at 20:25