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Take a function $f:\mathbb{R}\rightarrow(0,+\infty)$ non-decreasing and such that $\mathrm{lim\;inf}_{n\rightarrow+\infty}(f(n+1)-f(n))>0$ then $\mathrm{lim\;sup}_{x\rightarrow+\infty}\;\frac{f(x)}{x}>0$.

The only idea that I got is to prove by contradiction that there is a convergent subsequence of $(f(n+1)-f(n))$ to a limit $\leq0$. But I have no idea how to use this idea, could you help me please?

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Just use the definition of liminf directly: Choose $\delta$ strictly between $0$ and $\liminf(f(n+1)-f(n))$; then for some $N$ onwards each increase in $f$ is at least $\delta$. Therefore $f(n)\ge c+\delta(n-N)$ where $c=f(N)$.

Since $\frac{c+\delta(n-N)}{n}\to \delta$, the conclusion follows.

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  • $\begingroup$ i don't understand why $f(n)\geq c+\delta(n-N)$, could you explain me, please? $\endgroup$ – Mec Oct 18 '11 at 2:06
  • $\begingroup$ $f(N)=c$ and each time the argument increases by 1, the result must increase by at least $\delta$. If you want to be formal about it, use induction on $n-N$. $\endgroup$ – Henning Makholm Oct 18 '11 at 2:13
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This might be slightly off-topic, but this is special case of Stolz–Cesàro theorem in the following form:

Theorem. If $(a_n)$, $(b_n)$ are two real sequences such that $(b_n)$ positive, unbounded and strictly increasing, then $$\liminf \frac{a_{n+1}-a_n}{b_{n+1}-b_n} \le \liminf \frac{a_n}{b_n} \le \limsup \frac{a_n}{b_n} \le \limsup \frac{a_{n+1}-a_n}{b_{n+1}-b_n}.$$

If you use this result for $b_n=n$, then you get your result.


I am well aware of the fact, that this is an overkill in this case, but I could not resist - I have the feeling that this form of Stolz-Cesaro (including liminf and limsup) is less known than it deserves.

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To be precise, the answers of Henning Makholm and Martin Sleziak insists that $\displaystyle \limsup_{n\rightarrow \infty} \dfrac{f(n)}{n} >0$. Now, $\displaystyle \limsup_{x\rightarrow \infty} \dfrac{f(x)}{x} \geq \limsup_{n\rightarrow \infty} \dfrac{f(n)}{n}>0$, which is the conclusion what we want.

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