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 $(a_k)_{k \in \mathbb{N}} \in \ell^p(\mathbb{N})$?


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .