Proof of limits of sequences tending to infinity How can we prove this :
Prove that if $\quad\lim_{n\to+\infty} y_n = \lim_{n\to+\infty} z_n=+\infty$ 
then $\quad\lim_{n\to+\infty} v_n = \lim_{x\to+\infty} w_n=+\infty$ .
With  $$w_n=\frac{n}{\sum_{i=1}^n\frac{1}{z_i}},\quad v_n = \frac{y_1+y_2+....y_n}{n} $$.Can we solve this with "epsilon delta" proof? Or can we solve this with another way? I don't find anything to get started
 A: Consider the second sequence first. Take any $M>0$. Then by assumption $y_n>2M$ for all $n\geq n_0$ for some $n_0$. Moreover $\sum_{j=n_0}^n y_j\geq (n-n_0+1)2M\geq 2nM$ and $\sum_{j=1}^{n_0-1} y_j\geq -n_0 c$ with $\max_{1\leq j\leq n_0-1}{\vert y_j\vert}=:c$. Therefore $v_n\geq -\frac{n_0}{n}c+2M$. Now choose $n_1\geq n_0$ such that $\frac{n_0}{n}c\leq M$ for all $n\geq n_1$. Then $v_n\geq -M+2M=M$ for all $n>n_1$.
The first part  (and to some extend also the second) may be solved by arithmetic mean of a (convergent) sequence converges.
A: $y_n \to \infty \implies$ For every $\epsilon \gt 0 \;\;\exists N$ such that for all $n\gt N$, $|y_n|\gt \epsilon$ For sufficiently large $n$, We have, $v_n=
\frac{\sum_{i=0}^{i=n}y_i}{n}=\frac{\sum_{i=0}^{i=N}y_i+\sum_{i=N+1}^{i=n}y_i}{n}\gt \frac{\sum_{i=0}^{i=N}y_i}{n}+\frac{\epsilon (n-N)}{n}$ 
As $n\to \infty$, we have, $\lim v_n=\lim \frac{\sum_{i=0}^{i=n}y_i}{n}\ge \epsilon $  
Since, $\epsilon \gt 0$ is arbitrary, $\lim v_n$ tends to infinity. 
Hint for $w_n$: You just need to show that if $(1/z_n)\to 0$, then $(1/w_n)\to 0$. Proceed as above to prove this.
