Limit of the reciprocal of the mean harmonic Suppose $(x_n)$ is a convergent sequence such that $x_n>0 \forall n \in \mathbb{N}$.
Set $$y_n=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}, n\in\mathbb{N}$$
Prove $\lim_{n\to\infty}{x_n}=\lim_{n\to\infty}{y_n}$ <- How should I approach this?
Also prove that if $x_n$ diverges to $+\infty$, $(y_n)$ diverges to $+\infty$ <- I suppose I should be able to do this if I prove the above statement?
 A: Note that
$$
\frac{1}{y_n}=\frac{1}{n}\left(\frac{1}{x_1}+\cdots+\frac{1}{x_n}\right).
$$
So, as $x_n>0$, we have that
$$
\lim_{n\to\infty}\frac{1}{x_n}=\left\{\begin{array}{lll} \frac{1}{\lim_{n\to\infty}x_n} &
\text{if} & \lim_{n\to\infty}x_n>0, \\ \infty & \text{if} & \lim_{n\to\infty}x_n=0.
\end{array}\right.
$$
Finally, Cesaro Test provides that
$$
\frac{1}{y_n}\to \lim_{n\to\infty}\frac{1}{x_n},
$$
and likewise
$$
y_n\to \lim_{n\to\infty}x_n.
$$
A: If $x:=\lim x_n\ne0$ then $\frac1{x_n}$ converges to $\frac 1x$ (this also works for  $x=+\infty$ where we interprete $\frac1x$ as $0$) and we see that $\frac1{y_n}$ also converges to $\frac 1x$ by Cesàro. This is also true for $x=0$, i.e. when $\frac1{x_n}$ diverges to $+\infty$, then teh Cesàro mean $\frac1{y_n}$ must also diverge to $+\infty$.
A: For the second part: as
$$
\frac 1{x_n} \to 0
$$you can use the Cesaro theorem, which concludes that
$$
\frac 1y_n = \frac{\frac1{x_1}+ \dots +\frac1{x_n}}{n} \to 0
\\\implies y_n \to + \infty
$$(as $y_n \ge 0$ the limit is $>0$).
For the first part, you can do basically the same thing. 
