Two different results of calculation of the limit of $(2+4+\dots+2n)/n^2$

Question

I would like to know where is the problem when i calculated those two limit with the following ways : method (1):

$$\lim_{n\to \infty} \dfrac { 2+4+6+\cdots 2n} {n²} = \lim_{n\to \infty} \dfrac {n(n+1)}{n²} =1 $$( using arithmitic sequence sum ).

Method (2) : now by using operation over limit we get this: {n²}$$\lim_{n\to \infty} \dfrac { 2+4+6+\cdots 2n} {n²} = \lim_{n\to \infty} \dfrac {2}{n²}+\lim_{n\to\infty}\dfrac {4}{n²}+\cdots \lim_{n\to\infty}\dfrac {2n}{n²} =0+0+\cdots 0=0 $$ .

Then is there some one who show me where is the problem in second method ? Thank you for any help

How many terms are you adding here? $$\lim_{n\to \infty} \dfrac {2}{n²}+\lim_{n\to\infty}\dfrac {4}{n²}+\cdots \lim_{n\to\infty}\dfrac {2n}{n²}$$ — , Nov 05 '15 at 17:12
@zeraouliarafik I think Bungo meant that question as a question for you to reflect on. It's really at the core of the issue. Read mvw's answer for clarification. — Scounged, Nov 05 '15 at 17:16
Would you also agree $$\lim_{n\to \infty}\frac{n}{n}=\lim_{n\to \infty}\left(\frac{1}{n}\right)+\lim_{n\to \infty}\left(\frac{1}{n}\right)+\cdots+\lim_{n\to \infty}\left(\frac{1}{n}\right)=0+0+\cdots+0=0$$ — user236182, Nov 05 '15 at 18:41

mvw · Answer 1 · 2015-11-05T17:37:17.043

3

You are trying $$ \lim_{n\to\infty} \sum_{k=1}^n \frac{2k}{n^2} "=" \sum_{k=1}^n \lim_{n\to\infty} \frac{2k}{n^2} "=" \sum_{k=1}^n \lim_{m \to\infty} \frac{2k}{m^2} $$ but $n$ has to be constant within the summation scope. For the middle term it can not stay both fixed and be subject of a limit. Only the first and last term are well defined (and different).

edited Nov 05 '15 at 17:37

answered Nov 05 '15 at 17:14

mvw

34,562

score -1 · Answer 2 · answered Nov 05 '15 at 17:11

-1

in method 2 you are summing an infinite number of zeros: the result is undefined, not 0.

answered Nov 05 '15 at 17:11

Fabrice NEYRET

962

2

The problem occurs earlier than that; it's with mimicking "limit of sum is the sum of limits" in a situation when it doesn't apply. – Nov 05 '15 at 17:13
but 0 for n times is 0 – zeraoulia rafik Nov 05 '15 at 17:14
Yes! The mistake is realy there. – kmitov Nov 05 '15 at 17:14
No, that's not a mistake at all. $\lim_{n\to \infty}\sum_{i=1}^n 0=0$. – user236182 Nov 05 '15 at 18:54
it's not 0 n times, its small amount n times, tending towards 0 infinite times. – Fabrice NEYRET Nov 05 '15 at 20:15

Two different results of calculation of the limit of $(2+4+\dots+2n)/n^2$

2 Answers2