decresing sequence implies $\lim{n u_n} $ exists? if $(u_n)$ is a nonincreasing sequence of positive real
numbers; how to prove or disprove  that the sequence $(nu_n)$ admits a limit?
We know  two results
1- if $(u_n)$ is a  sequence of positive real
numbers such that $\sum_n a_n$ converges, and if $\lim\limits_{n \to \infty} n u_n$ exists, then $\lim\limits_{n \to \infty} n u_n = 0$
2- if $(u_n)$ is a nonincreasing sequence of positive real
numbers such that $\sum_n u_n$ converges, then $\lim_{n \rightarrow \infty} n u_n = 0$.
PS :  I corrected the usual notation for a sequence
Kenta S gave two valid examples in their comment: both $u_n = \frac1{\sqrt n}$ and $u_n=1$ have the property that $\lim_{n\to\infty} nu_n = \infty$. It's important to note that, despite the use of the equals sign in that notation, it indicates a specific way that $\lim_{n\to\infty} nu_n$ fails to exist. (Infinity is not a number.)
But other divergent behaviour is also possible. Consider for example
$$
u_n = 2^{-\lfloor \log_2 n\rfloor} = \min\bigg\{ \frac1{2^k}\colon \frac1{2^k} &gt; \frac1n \bigg\},
$$
so that $(u_n) = (1,\frac12,\frac12,\frac14,\frac14,\frac14,\frac14,\frac18,\dots)$. It's not hard to show that $\limsup_{n\to\infty} nu_n = 2$ but $\liminf_{n\to\infty} nu_n = 1$, and so $\lim_{n\to\infty} nu_n$ does not exist.
Let $y_n = n u_n$. Then $L = \lim_{n \rightarrow \infty} y_n$ exists iff $\forall \varepsilon \exists m_0 \forall m \geq m_0 \mbox{ s.t. }|y_m - L| &lt; \varepsilon$, i.e. for any small value, the $y_n$ beyond some point are all closer to $L$ than that value.
This means that $|n u_n - L| &lt; \varepsilon$, so $|u_n - \frac{L}{n}| &lt; \frac{\varepsilon}{n}$, or to rearrange it (and taking advantage of the fact that $u_n &gt; 0$), $\frac{L - \varepsilon}{n} &lt; u_n &lt; \frac{L + \varepsilon}{n}$.
This means that the limit can only exist if $u_n = o(n^{-1})$ , i.e. $u_n$ shrinks faster than $\frac{1}{n}$.