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Let $(a_k)_{k \in \mathbb{N}} \subset ]0,+\infty[$, and assume that $a_k \to 0$ as $k$ goes to $+\infty$, but $a_k \notin \ell^1(\mathbb{N})$.

It is easy to prove that we can extract a subsequence $(a_{k_n})_{n \in \mathbb{N}} \in \ell^1(\mathbb{N})$; for this it is enough to choose $k_n$ going to $+\infty$ "fast enough".

I wonder if the following implication is true: $(a_{k_n})_{n \in \mathbb{N}} \in \ell^1(\mathbb{N})$ $\Rightarrow$ $(\frac{1}{k_n})_{n \in \mathbb{N}} \in \ell^1(\mathbb{N})$ ?

I also wonder if there is a way to reverse this implication, by doing an additional summability assumption like $(a_k)_{k \in \mathbb{N}} \in \ell^p(\mathbb{N})$?

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The first implication is false. Here is a counter example. Define the following sequence (I include $0$ in the natural numbers $\mathbb{N}$),

$$\{a_k\}_{k\in\mathbb{N}} = \left\{1, 10^{-1} ,\frac{1}{2}, 10^{-2}, \frac{1}{3}, 10^{-3} ,\ldots\right\}=\begin{cases}\frac{1}{\frac{k}{2}+1} & k \mbox{ even}\\ 10^{-(\frac{k-1}{2}+1)} & k\mbox{ odd} \end{cases}$$

which is the sequence $\left\{\frac{1}{k+1}\right\}_{k\in\mathbb{N}}$ spliced with a geometric sequence with ratio $10^{-1}$. Obviously $a_k\to 0$ with $a_k> 0$ but we also have $\{a_k\}_{k\in\mathbb{N}}\not\in\ell^1(\mathbb{N})$ since our partial sum is lower bounded by a partial sum of the harmonic series.

Then we can form the subsequence $a_{k_n} = a_{2n+1} = 10^{-(n+1)}$ which is summable but for which the sequence $\left\{k_n^{-1}\right\}_{n\in\mathbb{N}}$ is not summable since,

$$\sum\limits_{n=0}^\infty \frac{1}{k_n} = \sum\limits_{n=0}^\infty \frac{1}{2n+1}=\infty.$$


As for the reverse implication... It is not true as stated, i.e. without additional assumptions on $\{a_k\}_{k\in\mathbb{N}}$, and I suspect it's not true even with an assumption like $\{a_k\}_{k\in\mathbb{N}}\in \ell^p(\mathbb{N})$.

To see that the reverse implication is not true consider the following scenario. You take $\{a_k\}_{k\in\mathbb{N}}$ to be the typical $\left\{\frac{1}{k+1}\right\}_{k\in\mathbb{N}}$ except that whenever $k$ is a square you replace $a_k$ by $\frac{1}{\sqrt{k}}$,

$$a_k = \begin{cases} \frac{1}{k} & \sqrt{k}\not\in\mathbb{N} \\ \frac{1}{\sqrt{k}} & \sqrt{k}\in\mathbb{N}\end{cases}$$

which gives something like $\{a_k\}_{k\in\mathbb{N}} = \left\{\color{red}{1}, \frac{1}{2}, \frac{1}{3}, \color{red}{\frac{1}{2}}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}, \color{red}{\frac{1}{3}}, \ldots\right\}$

This still satisfies the assumptions we've set forth, i.e. $a_k\to 0$, $a_k>0$, and $\{a_k\}_{k\in\mathbb{N}}\not\in\ell^1(\mathbb{N})$. The sequence $\left\{k_n^{-1}\right\}_{n\in\mathbb{N}} = \left\{\frac{1}{(n+1)^2}\right\}_{n\in\mathbb{N}}$ is obviously summable but these indices correspond exactly to the red subsequence $\{a_{k_n}\}_{n\in\mathbb{N}}$ which is now $\left\{\frac{1}{n+1}\right\}_{n\in\mathbb{N}}\not\in\ell^1(\mathbb{N})$.

edit: actually, the sequence I gave above is in $\ell^p(\mathbb{N})$ for all $p>1$ and so it is clear that stronger assumptions will be necessary if the converse is going to hold.

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