Show that if $\sum_na_n=\infty$ and $a_n\downarrow 0$ then $\sum\limits_n\min(a_{n},\frac{1}{n})=\infty$ 
Let $a_n$ be a sequence decreasing to $0$, and $\sum {{a_n} = \infty } $.
  Show that:$$\sum {\min \left( {{a_n},{1 \over n}} \right)}  = \infty $$

If there's $N_0$ such that $\forall n>N_0: \min(a_n, {1\over n}) = {1\over n}$ 
or $\forall n>N_0: \min(a_n, {1\over n}) = a_n$ then, the problem is trivial.  
Otherwise, let us define $b_n = \min(a_n, {1\over n})$ and two subsequences:
$$b_{n_k} = {1 \over n_k},\quad b_{n_l} = a_{n_l}$$
$$\sum {{b_n} = } \sum {{b_{{n_k}}}}  + \sum {{b_{{n_l}}}}  = \infty  + \infty $$
Is that right?
 A: 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 $\sum_{n=k}a^n=\infty$.
A: 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.
