Existence of $p_n$ such that $\lim_{n \to \infty} \frac{a_1+a_2+\dots+a_{p_{n+1}}}{a_1+a_2+\dots+a_{p_n}}=l$

Question

Let $0<a_n \leq 1$ for all $n \geq 1$ and $\sum_{k=1}^{\infty}a_k=\infty$. Let $l \in [1,\infty]$. Prove that there is a strictly increasing sequence of natural numbers $p_n$ such that $$\lim_{n \to \infty} \frac{a_1+a_2+\dots+a_{p_{n+1}}}{a_1+a_2+\dots+a_{p_n}}=l$$

I could only prove it for $l=1$. I considered $p_n$ to be the smallest number such that $$a_1+a_2+\dots+a_{p_n}>n \quad (*)$$ This $p_n$ exists because $\sum_{k=1}^{\infty}a_k=\infty$. Then we also have $$a_1+a_2+\dots+a_{p_n-1}\leq n \Rightarrow a_1+a_2+\dots+a_{p_n}\leq n+a_{p_n}\leq n+1$$ But , these mean that $$\lim_{n \to \infty}\frac{a_1+a_2+\dots+a_{p_n}}{n}=1$$ and this shows that $$\lim_{n \to \infty} \frac{a_1+a_2+\dots+a_{p_{n+1}}}{a_1+a_2+\dots+a_{p_n}}=\lim_{n \to \infty}\frac{a_1+a_2+\dots+a_{p_{n+1}}}{n+1}\cdot \frac{n}{a_1+a_2+\dots+a_{p_n}}\cdot \frac{n+1}{n}=1 $$ I feel that this way may lead to the conclusion, and I tried to replace the $n$ in the right hand side of $(*)$ with something related to $l$, but the final result always goes to $1$.

Answer 1

First consider the case $1 \le l < \infty$: Define $p_0 = 1$ and then (recursively) $p_{n+1}$ as the first integer such that $$ x_n = \frac{a_1+a_2+\cdots+a_{p_{n+1}}}{a_1+a_2+\cdots+a_{p_n}} > l \, , $$ this is always possible because $\sum_{k=1}^{\infty}a_k=\infty$. Then $p_{n+1} > p_n$ and $$ l \ge \frac{a_1+a_2+\cdots+a_{p_{n+1}-1}}{a_1+a_2+\cdots+a_{p_n}} = x_n - \frac{a_{p_{n+1}}}{a_1+a_2+\cdots+a_{p_n}} \ge x_n - \frac{1}{a_1+a_2+\cdots+a_{p_n}} $$ So we have $p_n \to \infty$, $$ 0 < x_n - l \le \frac{1}{a_1+a_2+\cdots+a_{p_n}} $$ and the right-hand side converges to zero.

In the case $l = \infty$ one can proceed similarly: Define $p_0 = 1$ and then (recursively) $p_{n+1}$ as the first integer such that $$ x_n = \frac{a_1+a_2+\cdots+a_{p_{n+1}}}{a_1+a_2+\cdots+a_{p_n}} > n \, . $$