How to show that if $\sum_na_n=\infty$ and $a_n\downarrow 0$ then $\sum\limits_n\min(a_{n},\frac{1}{n})=\infty$? Suppose $a_n\downarrow 0, \sum\limits_{n=1}^{\infty}a_n=+\infty,   b_n=\min\{a_n,1/n\}$. Prove that $\sum b_n $ diverges.
In fact, I have known that two positive divergent series $\sum a_n ~\sum b_n$,  $c_n=\min\{a_n,b_n\}, \sum c_n$ is not always divergent.
But I do not know why this above series is surely divergent. Sincerely thanks.
 A: 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 
A: 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.
