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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

14max
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2 Answers2

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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.

Jens Schwaiger
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$y_n \to \infty \implies$ 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.

Koro
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  • how can we assume this $v_n=\frac{\sum_{i=0}^{i=n}y_i}{n}\to \infty$ ? – 14max May 29 '20 at 07:00
  • how do we apply the limit compariso test in here? – 14max May 29 '20 at 07:14
  • Yeah ,I understand this definition but I don't get how can we assume that "$v_n$>$y_n$ " from this expression? – 14max May 29 '20 at 07:21
  • Note the first line in answer. $y_n \gt \epsilon$. So $\sum_{i=N+1}^{i=n}y_i\gt \epsilon (n-N)$. Define $u_n=\frac{\sum_{i=0}^{i=N}y_i}{n}+\frac{\epsilon (n-N)}{n}$. So we have $v_n\gt u_n$. Now what is $\lim u_n$ as $n \to \infty$? – Koro May 29 '20 at 07:24
  • And how about the second one?can you give me a hint to start it?just hint not a full solution this time ,I don't get used to deal with infinit limits – 14max May 29 '20 at 07:34
  • You may use the hint in answer. Just proceed as in case of $v_n$. Inequality will be $\lt $ instead of $\gt$. – Koro May 29 '20 at 07:39
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    The expected limit of $u_n$ is +$\infty$ right ?but how?one last question – 14max May 29 '20 at 07:43
  • I just realized that I made some mistakes in my answer. Thanks a lot for pointing out the mistakes. Indeed,as you have rightly pointed out $\lim u_n$ is not $\infty$ rather it is $\epsilon$. I have updated my answer now. Sorry for causing confusion. – Koro May 29 '20 at 08:49
  • "As $n\to \infty$, we have, $\lim v_n=\lim \frac{\sum_{i=0}^{i=n}y_i}{n}\ge \epsilon $ " ,so this means that $\frac{\sum_{i=0}^{i=N}y_i}{n}+\frac{\epsilon (n-N)}{n}$ >ϵ ? – 14max May 29 '20 at 08:52
  • I'm just new to these kinds of proof – 14max May 29 '20 at 08:58
  • No. If $q_n \gt a$ then we have $\lim q_n \ge a$. But not the other way round. For example, $\lim (-1/n)=0\ge 0$ but $(-1/n)\lt 0$ – Koro May 29 '20 at 08:59
  • "$n\to \infty$, we have, $\lim v_n=\lim \frac{\sum_{i=0}^{i=n}y_i}{n}\ge \epsilon $" I don't really get it then ? – 14max May 29 '20 at 09:07
  • Example: $(1+1/n) \gt 1$. Right? But $\lim (1+1/n)=1\ge 1$ In general, there is a result which says that if $x_n\gt y_n$ then $\lim x_n \ge \lim y_n$ if both the limits exist. – Koro May 29 '20 at 09:15
  • yeah ,I understand ,how can this tend to infinity then?$v_n$? – 14max May 29 '20 at 09:26
  • Because $\epsilon $is any arbitrary positive number. $\lim v_n $ is greater than any arbitrarily chosen positive $\implies $ the limit is infinity. If not, then let some positive number $l$ be lim ($v_n$), then choose $\epsilon $as $l+1$, we have a contradiction. – Koro May 29 '20 at 09:34