Asymptotic Analysis of a Sequence Problem: Convergence of $\log(n)a_n$ as $n\to+\infty$

Question

Analyzing the Convergence of $\log(n)a_n$ in a Sequence Problem

Consider two positive sequences $(a_n)_{n \geq 0}$ and $(b_n)_{n \geq 0}$ satisfying the following conditions:

• $(a_n)_{n \geq 0}$ is a decreasing sequence.
• $nb_n \longrightarrow 1$ as $n\longrightarrow +\infty$.
• For all $n \in \mathbb{N}$, $\sum_{k=0}^{n}a_kb_{n-k} = 1$.

The objective is to demonstrate that $\log(n)a_n \longrightarrow 1$ as $n\longrightarrow +\infty$.

In attempting a solution, I observed that $b_n \sim \frac{1}{n}$, implying $\sum b_n \sim \log(n)$ by the properties of the harmonic series. Seeking insights and solutions to address this gap in the analysis.

Answer 1

Proof that $\liminf_{n \to \infty } \log(n) a_n \ge 1$

Note that by @Martin proof we have that $a_n \to 0$ hence $\frac{1}{n}(\sum_{k=0}^{n-1}a_k) \to 0, n \to \infty$. Also $n b_n \to 1$ means $b_n \le 2/n, n \ge n_0$ so for all $\epsilon >0$ there is $n_{\epsilon}$ st $$\sum_{k=0}^{N-1}a_kb_{2N-k} \le 2\sum_{k=0}^{N-1} \frac{a_k}{2N-k} \le \frac{2}{N}(\sum_{k=0}^{N-1}a_k) \le \epsilon$$ for all $N \ge n_{\epsilon}$

Hence $1=\sum_{k=0}^{2N} a_kb_{2N-k}$, $a_N$ decreasing and $N \ge n_{\epsilon}$ implies $$a_{N}\sum_{k=N}^{2N}b_{2N-k}+ \epsilon \ge \sum_{k=N}^{2N}a_kb_{2N-k}+ \sum_{k=0}^{N-1}a_kb_{2N-k}=1$$

But now $\sum_{k=N}^{2N}b_{2N-k}=\sum_{k=0}^{N}b_k=B_N$ so for for $N \ge n_{\epsilon}$ we get $$a_NB_N \ge 1-\epsilon, N \ge n_{\epsilon}$$ hence $\frac{B_N}{\log N} (a_N \log N) \ge 1-\epsilon$ and since $\frac{B_N}{\log N} \to 1$ it follows that $\liminf_{N \to \infty} (a_N \log N) \ge 1-\epsilon$

Since $\epsilon >0$ was arbitrary we get $\liminf_{N \to \infty} (a_N \log N) \ge 1$ and we are done completing @Martin partial answer

Answer 2

A partial result.

Let $B_n = \sum_{k=0}^n b_k$. Then $$ \lim_{n \to \infty} \frac{B_n}{\log (n)} = \lim_{n \to \infty} \frac{b_n}{\log(n) - \log(n-1)} = \lim_{n \to \infty} n b_n \cdot \frac{1/n}{\log( 1+1/(n-1))} = 1 $$ by the Stolz–Cesàro theorem. $(a_n)$ is decreasing, so that $$ 1 = \sum_{k=0}^{n}a_kb_{n-k} \ge a_n B_n $$ and therefore $$ \limsup_{n \to \infty } \log(n) a_n \le \limsup_{n \to \infty } \frac{\log(n)}{B_n} = 1 \, . $$

It remains to show that $\liminf_{n \to \infty } \log(n) a_n \ge 1$.