Does $\lim \sup x_{n+1}-x_n=+\infty \implies \lim \dfrac{n}{x_n}=0$

If $(x_n)$ is a real sequence such that $\lim \sup x_{n+1}-x_n=+\infty$ , then must we have $\lim \dfrac{n}{x_n}=0$ ?

• $1, 2^1,2,2^2,3,2^3,\ldots$ is a counterexample. – David Mitra Feb 14 '14 at 13:25
• @DavidMitra: Do you mean $x_{2n-1}=n$ and $x_{2n}=2^{n}$ – user123733 Feb 14 '14 at 13:28
• Yes. Essentially you select the odd terms so that $\lim_n (n/x_n)=0$ fails. Then select even terms so that the limsup is infinite. – David Mitra Feb 14 '14 at 13:30
• @DavidMitra: Ok , got it. – user123733 Feb 14 '14 at 13:38

Some non-monotone examples were given in comments. But even if $(x_n)$ is additionally assumed to be increasing, the answer remains negative. For $n =1,2,3,\dots$ define $$x_{n^3}=n^2, \qquad x_{n^3+1}=n^2+n$$ and extend to the rest of indices by linear interpolation.

If you assume $\liminf (x_{n+1}-x_n)=+\infty$, then the conclusion is true (and easy to prove).

Let it be: $\lim_{x\to \infty } \, \frac{x}{f(x)}\neq 0$ and $\lim_{x\to \infty } \, |f(x)|=+\infty$ then:

$\lim_{x\to \infty } \, |f'(x)| \neq +\infty$ , if it is $+\infty$ then, by L'Hôpital's rule $\lim_{x\to \infty } \, \frac{x}{f(x)}=\lim_{x\to \infty } \, \frac{1}{f'(x)}=0$ which is in conflict with assumption that: $\lim_{x\to \infty } \, \frac{x}{f(x)}\neq 0$, so: $$\lim_{X\to \infty } \, \int_X^{X+1}f'(x)dx=\lim_{X\to \infty } \, f(X+1)-f(X)\neq\infty$$ $$(\lim_{x\to \infty } \, |f'(x)| \neq +\infty)$$

If $\lim_{x\to \infty } \, |f(x)|\neq \infty$ then we can't use L'Hôpital's rule,but situation is obvious and $\lim_{X\to \infty } \, f(X+1)-f(X)$ is also not $\infty$.

So, I've proved that if $\lim_{x\to \infty } \, \frac{x}{f(x)}\neq 0$ then $\lim_{X\to \infty } \, f(X+1)-f(X)\neq\infty$ which means if $\lim_{X\to \infty } \, |f(X+1)-f(X)|=+\infty$ then $\lim_{x\to \infty } \, \frac{x}{f(x)}=0$.

Note: There must exist $x_0$, that for any $x$ greater than $x_0$ , $f(x)$ has real values and derivative.

If $\lim_{x\to \infty } \, f'(x)$ exists and is equal to k then $k=\lim_{x\to \infty } \, \frac{f(x)}{x}=\lim_{x\to \infty } \, f(x+1)-f(x)$, which is useful and related with $O(n)$ functions.

Summary:

Theory that: $$\lim\sup x_{n+1}-x_n=+\infty\ \rightarrow \lim \frac{n}{x_n}=0$$ is false if we consider all sequences with real values:

Sample:

Let's consider: $$x(n) = (-1)^n\cdot\sqrt{n}$$ $\lim \sup x(n+1)-x(n) = +\infty$ , but $\lim \frac{n}{x(n)}$ is not zero. But theory is true for large family of sequences $f(n)$ : $f(x)$ has real values and derivative for any real $x$ greater than some real value $x_0$ related with considered function.