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If the monotonic real sequence $x_n \to 0$,then $$n(x_{n+1} -x_n) \to 0$$

  1. In addition to some obvious examples(such as $\frac{1}{n^a},a>0$), I also verified some examples of sequences that converge very slowly.

    For example: $$n\Big(\frac{1}{\log(\log(n+1))} - \frac{1}{\log(\log(n))}\Big)\sim -\frac{1}{\log(n)\log(\log(n))^2}$$

  2. I already know that the monotonicity condition is necessary, otherwise there would be a counterexample like this： $$n\Big(\frac{\sin(n+1)}{n+1} - \frac{\sin(n)}{n}\Big) \not \to 0$$

Henry
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    No it is not a counterexample,$n(\frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n}}) \sim -\frac{1}{2\sqrt{n}} \to 0$ – Henry Feb 01 '23 at 05:57
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    @MartinR: (retracted, should just give the link) – metamorphy Feb 01 '23 at 06:00
  • I think your expression for the meaning of $n(x_{n+1} -x_n) \not\to 0$ isn't right. It should be $\exists \varepsilon_0>0 s.t. \forall N> 0 \exists n>N s.t. n(x_{n+1} -x_n) \geq \varepsilon_0$. In other words, only some infinite subsequence needs to violate the bound.. – Alex K Feb 01 '23 at 06:06
  • @Alex K You are right,I found the problem in my proof. – Henry Feb 01 '23 at 06:22
  • Your examples $x_n = 1/n^\alpha$ and $x_n = 1/\log(\log(n))$ are of the form $x_n = f(n)$ with a convex decreasing function $f$. In that case $n(x_{n+1} -x_n) \to 0$ is true, I have added that as an addendum to my answer. – Martin R Feb 01 '23 at 08:15

2 Answers2

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Counterexamples can be constructed as follows: Let $(b_n)$ be any sequence of non-negative real numbers such that

  • $n b_n = 1$ for infinitely many $n$, and
  • The series $\sum_{n=1}^\infty b_n$ is convergent.

Then $$ x_n = \sum_{k=n}^\infty b_k $$ decreases to zero, and $$ n(x_{n+1}-x_n) = -n b_n = -1 $$ for infinitely many $n$.

Such sequences $(b_n)$ exist, for example $$ b_n = \begin{cases} 1/n & \text{ if $n = k^2$ for some positive integer $k$} \\ 0 & \text{otherwise.} \end{cases} $$

Addendum: If $(x_n)$ is a convex decreasing sequence converging to zero, i.e. if additionally $$ x_{n+1} \le \frac 12 \bigl(x_{n}+ x_{n+2}\bigr) $$ for all $n$ then $$ 0 \le x_{n+1}-x_{n+2} \le x_{n} - x_{n+1} $$ and therefore $$ 0 \le n \bigl(x_{n} - x_{n+1}\bigr) \le 2 \sum_{k=\lfloor n/2 \rfloor}^n \bigl(x_k - x_{k+1} \bigr) = 2 \bigl(x_{\lfloor n/2 \rfloor} - x_{n+1}\bigr) \to 0 $$

That explains why $n \bigl(x_{n} - x_{n+1}\bigr) \to 0$ in your test cases.

(This corresponds to the fact that if $\sum_{n=1}^\infty a_n$ is a convergent series with $(a_n)$ positive and decreasing then $n a_n \to 0$, see for example here.)

Martin R
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For $n\ge1$, let $x_n:=2^{-\lceil\log_2n\rceil}$ (i.e., $x_n=2^{-k}$ if $2^{k-1}<n\le 2^k$ for some $k\ge0$). Then $x_n\searrow0$ and, $$n(x_{n+1}-x_n)=\begin{cases}-\frac12,&\text{if $n$ is a power of $2$,}\\0,&\text{otherwise}.\end{cases}$$

nejimban
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