0

I'm stuck in this, I need some hint:

Prove that if a given sequence $z_n$ with $n\ge1$ converges to a limit $L$ then $w_n=\frac{n}{\sum_{i=1}^n\frac{1}{z_i}}$ converges to the same limit.

I tried a lot of things but it leads me nowhere.
I just need hints not full solution.

user376343
  • 8,311
14max
  • 37

2 Answers2

1

I assume that $L\in\mathbb{R}, L\neq 0.$

HINT $$z_n\to L \iff \frac{1}{z_n}\to \frac{1}{L}.$$

Therefore, ${1\over z_n}=\frac{1}{L}+\delta_n$ with $\delta_n\to 0.$

It suffices to realize that if $\delta_n\to 0$ then also ${1\over n} \sum\limits_{i=1}^n \delta_i \to 0.$
Indeed, $$w_n=\frac{n}{\sum\limits_{i=1}^n\frac{1}{z_i}}=\frac{1}{\frac{1}{n}\left({n\over L}+\sum\limits_{i=1}^n \delta_i\right)}$$

Can you finish it now?

user376343
  • 8,311
  • I don't really understand the use of $\delta$ in here.Especially in "$\frac{1}{n}$$\sum_{i=1}^n \delta_i$" – 14max May 22 '20 at 13:36
  • There was a typo. Try to manipulate $w_n$ now. – user376343 May 22 '20 at 13:39
  • I finally proved it with the method in the answer below,though I want understand your method also.I don't really understand your method,can you it be more clearer?I'm still a beginner at proof – 14max May 22 '20 at 15:04
  • You were asking for hints, not a solution. Now I see that I should have to complete the proof. I am adding now the computation. – user376343 May 22 '20 at 17:09
  • Yeah ,thank you so much :) .It makes a lot of sense now :) – 14max May 23 '20 at 08:59
1

Result cauhcy first theorem on limits : if sequence { $ a_n$} $\rightarrow l$ as $n\rightarrow \infty$

Then the sequence {$x_n$} also converges to $l$ where $$ x_n = \frac{a_1+a_2+....a_n}{n} $$ To your question If $\lim_{n\to\infty}z_n =L(\neq 0)$ $$\lim_{n\to\infty}\frac{1}{z_n }=\frac{1}{L}$$

Now take $a_n =\frac{1}{z_n}$ and $x_n = \frac{1}{w_n} $you will get your answer

ਮੈਥ
  • 426