Proving one limit is equal to another

Let ($x_s$) be a convergent sequence, where $x_s>0$ for all s, and $y_s$ be a sequence such that

$$\displaystyle \Large y_s=\frac{s}{\frac{1}{x_1}+...+\frac{1}{x_s}}$$

Prove $\displaystyle \lim x_s$=$\lim y_s$

So I know things like $x_s$ is bounded, and that it's Cauchy, but I can't think of anything to solve it. Any help is appreciated.

Solution 1:

If the sequence $(x_{s})_{s \in \mathbb{N}}$ converges to some $\ell > 0$ then :

The Cesaro theorem states that, since the sequence $\displaystyle \Big( \frac{1}{x_s} \Big)_{s \in \mathbb{N}}$ converges to $\displaystyle \frac{1}{\ell}$, we have :

$$ \frac{1}{s} \sum_{k=1}^{s} \frac{1}{x_k} \, \mathop{\longrightarrow} \limits_{s \to +\infty} \, \frac{1}{\ell}. $$

As a consequence,

$$ y_{s} = \Big( \frac{1}{s} \sum_{k=1}^{s} \frac{1}{x_k} \Big)^{-1} \, \mathop{\longrightarrow} \limits_{s \to +\infty} \, \ell = \lim \limits_{s \to +\infty} x_{s}. $$

If the sequence $(x_s)_{s \in \mathbb{N}}$ converges to $0$, the sequence $\displaystyle \Big( \frac{1}{x_s} \Big)_{s \in \mathbb{N}}$ converges to $+\infty$ and Cesaro again allows us to state that $\displaystyle \frac{1}{s} \sum_{k=1}^{s} \frac{1}{x_k} \, \mathop{\longrightarrow} \limits_{s \to +\infty} \, +\infty$. Therefore : $y_{s} \, \mathop{\longrightarrow} \limits_{s \to +\infty} 0$.