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


My  idea: 
$$b_{n}=\min(a_{n},\tfrac{1}{n})\Longrightarrow b_{n}\le a_{n},b_{n}\le\tfrac{1}{n}$$
then I don't know how to go on. 
 A: The desired conclusion in immediate  if we prove the following:

${\bf Proposition.}$ If $\{a_n\}$ is a positive decreasing sequece and if $~\sum_{n=1}^\infty\min(a_n,\frac{1}{n})$ is  convergent, then   $\sum_{n=1}^\infty a_n $ is also convergent.

Since $\sum_{n=1}^\infty\min(a_n,\frac{1}{n})$ is convergent, there is some $n_0$ such that
$$
\forall\, n\geq n_0,\quad\sum_{k=n+1}^{2n}\min\left(a_{2k},\frac{1}{2k}\right)<\frac{1}{2}\tag{1}
$$
Now, clearly $\min(a_{2n},\frac{1}{2n})\leq \min(a_k,\frac{1}{k})$ for $k\geq n$, 
So, $(1)$ implies that $n\min(a_{2n},\frac{1}{2n})<\frac{1}{2}$ for all $n\geq n_0$, or equivalently,
$$
\forall\, n\geq n_0,\quad  \min\left(na_{2n},\frac{1}{2}\right)<\frac{1}{2}\tag{2}
$$
But this shows that the minimum in the above inequality must be the term $na_{2n}$, therefore, $a_{2n} <\frac{1}{2n}$ for $n\geq n_0$. Let us write this as follows:
$$
\forall\, n\geq n_0,\quad   a_{2n} =\min\left(a_{2n},\frac{1}{2n}\right)\tag{3}
$$
Now the assumption that $\sum_{n=1}^\infty\min(a_n,\frac{1}{n})$ is convergent, implies the convergence of $\sum_{n=1}^\infty\min(a_{2n},\frac{1}{2n})$, and by $(3)$ it implies  the convergence of $\sum_{n=1}^\infty a_{2n}$, and since $a_{2n+1}\leq a_{2n}$ this implies also the convergence of $\sum_{n=1}^\infty a_{2n+1}$, and the announced conclusion follows.$\qquad\square$
