TITLE: Convergence of product of sum, and sum of reciprocals implies equality.
QUESTION [5 upvotes]: A friend claims this is true (and also claims he has a proof).
You are given the sequence $a_1, a_2, \dots, a_n, \dots$ which $a_n \gt 0$ and $p_n$ is defined as:
$$ p_n = (a_1 + a_2 + \dots + a_n)\left(\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}\right)$$
If $$\lim_{n \to \infty} \dfrac{p_n}{n^2}$$ exists (and is finite), then $a_i = a_j$ for any $i,j$.
This seems quite surprising. Is this really true?
I know we can show that the limit is $\ge 1$, by using the arithmetic and geometric mean inequality to show that $p_n \ge n^2$, but that seems to lead nowhere.
(Friend says the source is some forum he read somewhere, and does not remember).

REPLY [2 votes]: Assume that $a_n \to \alpha$ for some $\alpha \neq 0$. Then we always have
$$ \lim_{n\to\infty} \frac{p_n}{n^2} = 1. $$
This follows from the following Abelian theorem:

Theorem. If $x_n \to x$, then $\frac{1}{n}(x_1 + \cdots x_n) \to x$.

This shows that $(a_n)$ need not be eventually constant, since there are many convergent sequences which is not eventually constant.