Decreasing $\{a_n\}$ with divergent sum such that $\sum_{n=1}^\infty\min\{\frac1n,a_n\}$ converges

Question

Find a decreasing sequence $\{a_n\}$ convergent to $0$ with divergent series such that $\sum_{n=1}^\infty\min\{\frac1n,a_n\}$ converges.

This question is inspired by a contest question with harmonic series replaced by series of $\{\frac1{n\log n}\}$. The solution to this is not hard, first partition the series into blocks such that each block sum to $1+o(1)$. Then cut each block into half that one part corresponds to the major $1$ in the sum and another much larger part corresponds to $o(1)$ (In fact to a term in convergent series). Then we are led to replace the first part with smaller term that have convergent sum.

I believe that same idea can be applied to any decreasing series (limit is $0$) with order representable as an algebraic function like $\log\log n$ above. However, I cannot find a way to modify the method for harmonic series. Of course, any other method is perfectly fine.

New idea to add:

If $\{a_n\}$ is such a series, let $b_n=\min\{a_n,\frac1n\}$, then both $a_n$ and $\frac1n$ must appear infinitely often in $b_n$. Let $c_i$ be indices such that $b_n=\frac1n$.

We modify $\{a_n\}$ a bit without concerning whether the new series diverge or not. If $c_i\lt n\lt c_{i+1}$, then we have $a_n\ge a_{c_{i+1}}\ge \frac1{c_{i+1}}$. So we let $A_n=\frac1{c_{i+1}}$ then.

Let $B_n=\min\{A_n,\frac1n\}$, its sum has same order as $\sum\frac{c_{i+1}-c_i}{c_{i+1}}=\sum1-\frac{c_i}{c_{i+1}}$.

If $\{b_n\}$ form a convergent series, so is $\{B_n\}$, then the sum above have order $O(1)$, i.e. each $\frac{c_i}{c_{i+1}}=1+o(1)$.

So maybe we have to find an infinite product of some $d_n$ such that $\prod d_n=0$ and each has order $1+o(1)$ and in fact the remainder term there forms a convergent sum.

Idea added (2):

Then we are actually finding series $\{d_n\}$ such that $1\gt d_n\gt0$ and $\sum(1-d_n)$ converges and $\prod d_n=0$. However a theorem in infinite product shows that $\sum(1-d_n)$ converges absolutely iff $\sum\log{(1-d_n)}$ converges absolutely, then we have the infinite product converges, hence $\neq0$, contradiction.

Answer 1

This is not possible with the harmonic series.

Since $b_n = \min \: \bigl\{ \frac{1}{n}, a_n\bigr\}$ is monotonic, Cauchy's condensation test tells us that $\sum b_n$ converges if and only if $$\sum_{k = 1}^{\infty} 2^k \cdot b_{2^k}$$ converges. Suppose that $\sum b_n$ converges. Then the number of $k$ with $b_{2^k} = \frac{1}{2^k}$ must be finite, from which it follows that $$\sum_{k = 1}^{\infty} 2^k\cdot a_{2^k}$$ converges, and therefore $\sum a_n$ converges too.