Q:

series convergence imply nx approaches 0

Proof: For a decreasing sequence of positive reals, show that if the sum converges, then $nx_n \to 0$ but the converse is not true
The first part I just assumed a positive limit the series converge to, but the converse I’m failing to think of a counter example
A counter example could be $x_n=1/(n  \ln n )$
But if $x_n$ converges, does $(n \ln n)x_n\to 0$?

A:

We can use condensation test. Assuming we already know that if $\sum\limits_{n=1}^\infty x_n$ converges then $x_n = o(\frac{1}{n})$, we have
If $\sum\limits_{n=1}^\infty x_n$ converges then $\sum\limits_{n = 1}^\infty 2^n x_{2^n}$ converges. Then $2^n  x_{2^n} = o(\frac{1}{n})$. Substitute $k = 2^n$ we have $x_k = o(\frac{1}{k \log k})$ (for $k$ been power of two). We need to prove that $x_n = o(\frac{1}{n \log n})$ for arbitrary $n$ - ie $\forall \varepsilon > 0 \exists N_0 \forall n > N_0\colon x_n < \frac{\varepsilon}{n \log n}$. Let us use that $x_k = o(\frac{1}{k \log k})$ implies $x_k = o(\frac{1}{k (\log k + 1)})$
Let us take $N_0$ s.t. for any $k$ - power of $2$ greater than $N_0$ we have $x_k < \frac{\varepsilon / 2}{k (\log k + 1)}$. Now let $n$ be arbitrary integer greater than $N_0$ and $k$ be the greatest power of $2$ not greater then $n$ (so $k \leqslant n < 2k$). We have $x_n \leqslant x_k < \frac{\varepsilon / 2}{k (\log k + 1)} \leqslant \frac{\varepsilon / 2}{\frac{n}{2}(\log \frac{n}{2} + 1)} = \frac{\varepsilon}{n \log n}$.
We can repeat this process to get $x_n = o(\frac{1}{n\cdot \log n\cdot \log \log n\cdot \ldots\cdot\log \log \ldots \log n})$ for arbitrary number of logarithms.

