Is it true that $nx_n>(n+1)x_{n+1}$?

Question

Suppose that $\{x_n:n\ge1\}$ is a positive real sequence such that $x_n>x_{n+1}$ for each $n\ge1$ and $\sum_{n=1}^\infty x_n<\infty$. It follows that $nx_n\to0$ as $n\to\infty$.

Is it true that $nx_n>(n+1)x_{n+1}$ for large values of $n$?

It is straightforward to construct an example where this is the case (for instance, $x_n=n^{-2}$). Also, it does not necessarily hold for each $n$ (take, for example, $x_n=0.99^n$). I am struggling to construct a counterexample, but I nonetheless think that the answer is negative. Is it possible to construct an example, where the inequality does not hold for infinitely many values of $n$?

Any help is much appreciated!

Answer 1

What about this one?

$$\displaystyle x_n=\begin{cases} \frac{1}{n^2}&n\in 2\mathbb N\\ \frac{1}{n(n+1)}& n\in 2\mathbb N+1\end{cases} \\$$

Answer 2

It is not possible that the inequality holds only for finitely many $n$. Assume that, for all $n\geq N$,

$$nx_n \leq (n+1)x_{n+1}.$$

Then, as the sequence $a_n=nx_n$ is increasing along $n\geq N$, we have that

$$Nx_N \leq nx_n$$

for all $n\geq N$. So

$$\frac{Nx_N}{n} \leq x_n,$$

so $\sum_{n=N}^{\infty} x_n$ diverges by the comparison test.

For an example where both the inequality is false for infinitely many values of $n$ (and holds for infinitely many values of $n$), see Michael Biro's answer.