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

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.

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.

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.
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.