Let $(a_n)_n$ be a sequence of positive real numbers such that $a_n\leq1$ for all $n\in\mathbb{N}$, and suppose that $\displaystyle\lim_{n\to\infty}a_n=0$. By the Stolz–Cesàro theorem we know that $\displaystyle\lim_{n\to\infty}\left[\prod_{n=1}^na_n\right]^\frac{1}{n}=0$ $\quad(\star)$.

For each $n\in\mathbb{N}$, let $S_n$ be a random subset of $\{1,2,\dots,n\}$ and let $c_n$ be the cardinality of $S_n$. Put $$b_n=\left[a_{n+1}\left(\prod_{k\in S_n}a_k\right)\right]^\frac{1}{1+c_n}.$$

Question: Is $\displaystyle\lim_{n\to\infty}b_n=0$?

If I try to imagine cases, I only can think of $\displaystyle\lim_{n\to\infty}b_n=0$, but I don't know how to prove this.

For example, if $S_n\subset S_{n+1}$ (proper subset) for all $n\in\mathbb{N}$, then the factors that define $b_n$ form a sequence that converges to zero (subsequence of $(a_n)_n$), and therefore $\displaystyle\lim_{n\to\infty}b_n=0$ by $(\star)$.

Any insight for the general case would be appreciated.


Yes, this does hold.

Fix $\varepsilon \in (0, 1)$. Note that there are only finitely many $a_n$ such that $a_n \ge \varepsilon$; let $K$ be the number of such sequence terms (note that this depends on $\varepsilon$, not $n$). Since $a_n \to 0$, there exists some $N$ such that $$n > N \implies a_n < \varepsilon^{K+1}.$$

For each $n$, define \begin{align*} U_n &= \{n \in S_n : a_n \ge \varepsilon\} \\ L_n &= \{n \in S_n : a_n < \varepsilon\} = S_n \setminus U_n. \end{align*}

Please note that $|U_n| \le K$ for all $n$. Then, \begin{align*} n > N &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{K+1} \prod_{k\in S_n}a_k\\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{K+1} \prod_{k\in U_n}a_k \prod_{k\in L_n}a_k \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{|U_n| + 1} \prod_{k\in U_n}a_k \prod_{k\in L_n}a_k \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon \prod_{k\in U_n}\varepsilon a_k \prod_{k\in L_n}a_k \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon \prod_{k\in U_n}\varepsilon \prod_{k\in L_n}\varepsilon \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{|S_n|+1} \\ &\implies \left(a_{n+1}\prod_{k\in S_n}a_k\right)^{\frac{1}{|S_n|+1}} < \varepsilon. \end{align*}

  • 3
    $\begingroup$ This is beautiful! $\endgroup$ – Chilote Mar 20 at 5:34

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