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

There is a theorem (I believe called "Cauchy condensation test") that says that a series of positive decreasing terms $\sum_n s_n$ converges if and only if $\sum_k 2^k s_{2^k}$ converges. For your series you get that the series converges if and only if $\sum_k \min (2^k a_{2^k}, 1)$ converges. It is clear that this series converges if and only if $\sum_k 2^k a_{2^k}$ converges, which is if and only if $\sum_n a_n$ converges. So if $\sum_n a_n$ diverges then your series diverges.

Consider the sequence $b_n:=\min(a_n,\frac{1}{n})$. Note that it is non-negative and non-increasing: $$ 0\leq b_{n+1}\leq a_n\quad\hbox{and}\quad 0\leq b_{n+1}\leq \frac1n. $$ Thus, one can apply the Cauchy condensation to prove the divergence of $\sum b_n$. Now, we want to show that $$ \sum_{n=1}^\infty2^n\min\left(a_{2^n},\frac1{2^n}\right)=\infty. $$ Consider the following two sets: $$ A=\{n\in{\bf N}:a_{2^n}\geq\frac{1}{2^n}\},\quad B:=\{n\in{\bf N}:a_{2^n}<\frac{1}{2^n}\}. $$ If $A$ is infinite, then $$ \sum_{n=1}^\infty2^n\min\left(a_{2^n},\frac1{2^n}\right)\geq \sum_{n\in A}1=\infty. $$ If $A$ is finite, then there exists some positive integer $k$, such that $n\in B$ for all $n\geq k$. It follows that $$ \sum_{n=1}^\infty2^n\min\left(a_{2^n},\frac1{2^n}\right)\geq \sum_{n=k}^\infty2^na_{2^n}=\infty $$ where for the last equality we use the Cauchy condensation test and the fact that $\sum_{n=k}a^n=\infty$.