Sequence of real numbers (sequences)

Question

Let ${x_n,n\geq 1}$ be a sequence of real numbers in $(0,\infty)$ such that $\sup\{\frac{1}{x_{n}}\sum_{j=1}^{n}\frac{x_{j}}{j}:n\in\mathbb{N}\}<\infty$, Can it be concluded that $\lim_{n\rightarrow\infty}x_{n}=\infty$. Give a proof or else give a suitable counterexample.

Answer 1

We are given that there is an $L$ so that $$ \frac1{x_n}\sum_{k=1}^n\frac{x_k}k\le L\tag{1} $$ Inequality $(1)$ implies $$ x_n\ge\frac1L\sum_{k=1}^{n-1}\frac{x_k}k\tag{2} $$ For $n\ge2$, $(2)$ says $$ x_n\ge\frac{x_1}L\tag{3} $$ Plugging $(3)$ into $(2)$ yields $$ \bbox[5px,border:2px solid #C0A000]{x_n\ge\frac{x_1}{L^2}H_{n-1}}\tag{4} $$

Answer 2

Let $a_n=\frac {1}{x_n} \sum_{j=1}^{n} \frac {x_j}{j}.$

Then $\lim_{n \to \infty} a_n \le m \lt \infty$. ($\because \sup\{\frac{1}{x_{n}}\sum_{j=1}^{n}\frac{x_{j}}{j}:n\in\mathbb{N}\}=m<\infty$).

Suppose $\lim_{n \to \infty} x_n = l \lt \infty$ and $l \neq 0$.

Then $\frac {1}{l} \sum_{j=1}^{\infty} \frac {x_j}{j} \le m$.

$\Rightarrow \sum_{j=1}^{\infty} \frac {x_j}{j} \le ml$.

If $x=\inf \{x_n : n \in \Bbb N \}$ (note that $x \neq 0$), then $x \sum_{j=1}^{\infty} \frac 1{j} \le ml \; \Rightarrow \; \sum_{j=1}^{\infty} \frac 1{j} \le \frac {ml}{x}.$ Which is not possible...........($1$)

If $l = 0$, then we have $\lim_{n \to \infty} a_n = \dfrac {\lim_{n \to \infty} \sum_{j=1}^{n} \frac {x_j}{j}}{\lim_{n \to \infty} x_n} $. Now as $n \to \infty$, either $\sum_{j=1}^n \frac {x_j}{j}$ converges to a non-zero positive number or diverges. Thus in both cases, $a_n \to \infty$. Which is contradiction..........($2$)

Also if we suppose that $x_n$ is divergent, $\sup \{x_n : n \in \Bbb N\}=k \lt \infty$ and $x_n$ doesn't have a subsequence converging to zero, then $\frac 1k \sum_{j=1}^{\infty} \frac {x_j}{j} \le \lim_{n \to \infty} \frac {1}{x_n} \sum_{j=1}^n \frac {x_j}{j} \le m \Rightarrow \sum_{j=1}^{\infty} \frac {x_j}{j} \le mk$. Again let $x=\inf \{x_n : n \in \Bbb N\}$, then as in ($1$), we get contradiction...........($3$)

If $x_n$ is divergent, $\sup \{x_n : n \in \Bbb N\} \lt \infty$, and $x_n$ has a subsequence $s_n$ converging to zero :

Let $b_n$ be subsequence of $a_n$ with having terms of $s_n$ only. Then $\lim_{n \to \infty} b_n \le \lim_{n \to \infty} a_n \le m$. Going by same way as in ($2$), $\lim_{n \to \infty} b_n = \dfrac {\lim_{n \to \infty} \sum_{j(n)} \frac {s_{j(n)}}{{j(n)}}}{\lim_{n \to \infty} s_n}$. Now as $n \to \infty$, either numerator diverges or converges to a non-zero positive number. Thus, in both of these cases $b_n \to \infty$. Which is a contradiction..........($4$).