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

Question

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

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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.

Answer 1

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.$\qquad\square$