Is this true ?If $x_n \to 0$,then $n(x_{n+1} - x_{n}) \to 0$

Question

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

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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： $$n\Big(\frac{\sin(n+1)}{n+1} - \frac{\sin(n)}{n}\Big) \not \to 0$$

Answer 1

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 $$ b_n = \begin{cases} 1/n & \text{ if $n = k^2$ for some positive integer $k$} \\ 0 & \text{otherwise.} \end{cases} $$

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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.)

Answer 2

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, $$n(x_{n+1}-x_n)=\begin{cases}-\frac12,&\text{if $n$ is a power of $2$,}\\0,&\text{otherwise}.\end{cases}$$