construct a convergent positive series which $a_{n_k}\geq \frac{1}{n_k}$ How to construct a convergent positive series $\sum\limits_{n=1}^{\infty}a_n$, which has infinity terms satisfy $a_{n_k}\geq \frac{1}{n_k}$.
I constucted a series: $1,\dfrac{1}{4},\dfrac{1}{4},\dfrac{1}{4},\dfrac{1}{9},\dfrac{1}{9},\dfrac{1}{9},\dfrac{1}{9},\dfrac{1}{9}\cdots$, but I am not sure whether it's true.
If I'm wrong, please give me a correct example, otherwise help me prove it. Thanks!

Your series diverges because we can find the harmonic series inside it:
$$
   1 + \frac14 + \underbrace{\frac14 + \frac14}_{1/2} + \frac19 + \frac19 + \underbrace{\frac19 + \frac19 + \frac19}_{1/3} + \frac1{16} + \frac1{16} + \frac1{16} + \underbrace{\frac1{16} + \frac1{16} + \frac1{16} + \frac1{16}}_{1/4} + \dots
$$
In general, if the terms of the series have to be non-increasing, then there is no solution.
If they don't, then morally speaking we can just modify your example to take
$$
   1 + 0 + 0 + \frac14 + 0 + 0 + 0 + 0 + \frac19 + 0 + 0 + 0 + 0 + 0 + 0 + \frac1{16} + \dots
$$
This violates positivity, but we can replace the $0$ terms with something arbitrarily small (such as the terms of any convergent series).
