How to show that if $\sum_na_n=\infty$ and $a_n\downarrow 0$ then $\sum\limits_n\min(a_{n},\frac{1}{n})=\infty$?

Hint: Use Cauchy condensation test $\displaystyle\sum_{n=1}^{\infty}a_{n}$ convergent,if and only if $\sum_{n=1}^{\infty}2^na_{2^n}$ convergent

so we only show that $$\sum_{n=1}^{\infty}2^n\min\left(a_{2^n},\dfrac{1}{2^n}\right)$$ divergent

if some postive integer $n$ such $a_{2^n}\ge\dfrac{1}{2^n}$,then $$\sum_{n=1}^{\infty}2^n\min\left(a_{2^n},\dfrac{1}{2^n}\right)=\sum_{n=1}^{\infty}1$$ divergent

if some postive integer $n$ such $a_{2^n}<\dfrac{1}{2^n}$,then $$\sum_{n=1}^{\infty}2^n\min\left(a_{2^n},\dfrac{1}{2^n}\right)=\sum_{n=1}^{\infty}2^na_{2^n}$$ is divergent,because $\sum_{n=1}^{\infty}a_{n}$ divergent,then you can use Cauchy condensation test

Suppose instead that $\displaystyle\sum_{n=1}^{\infty}b_n$ converges.$\;$ Then $\displaystyle b_n\downarrow0\;$ (since $a_n\downarrow 0$ and $\frac{1}{n}\downarrow0$),

so $\displaystyle\lim_{n\to\infty}nb_n=0$ (as in the links shown below);

so $b_n=a_n$ for $n\ge N$ (for some $N\in\mathbb{N}$) and therefore $\displaystyle\sum_{n=N}^{\infty}b_n=\displaystyle\sum_{n=N}^{\infty}a_n$ diverges.

This gives a contradiction, so $\displaystyle\sum_{n=1}^{\infty}b_n$ diverges.