Question

A chicken manages to prove that: \[
\sum_{i=1}^{n}0 = 0
\]where \(n\in \mathbb{N}\).
She then continues by saying: \[
\lim_{n\to \infty}\sum_{i=1}^{n}0 = 0
\]In other words: \[
\underbrace{0+0+\dots+0}_{\infty} = 0
\]Is Mrs. Chicken correct? If not, what is the fallacy?

Answer 1

@anyone_with_a_brain

Answer 2

Woah , An SS 92 with a question

Answer 3

This is the hardest math I have ever seen. What math is this?

Answer 4

it's summation

Answer 5

Her proof is as follows: \[
\sum_1^10=0
\]So \(p(1)\) is true.
\[
\sum_1^{n+1}0=0+\sum_1^{n}0=0
\]So \(p(k)\implies p(k+1)\).

Answer 6

This question is like something satellite would ask :P

Answer 7

this question is like the key for immortality

Answer 8

ALL right all SS 90 below , stop bombarding this question with unnecessary comments , let the pros do their work :D

Answer 9

I already know the answer, it's just an interesting fact that is often overlooked.

Answer 10

@wio , are u in a university or at school ?

Answer 11

when you say lim n->infinity, you r just considering very large values of n
so, when very large numbers of 0's are added, it has to give infinity

Answer 12

has to give 0, i mean***

Answer 13

Do you agree with the chicken?

Answer 14

hartnn has a point , no matte how many times u add zero , it would still result in zero , Guess the chicken has brains

Answer 15

yeah...

Answer 16

What if the chicken says: \[
\underbrace{0+0+\dots+0}_\infty=\infty \times 0=0
\]

Answer 17

isn't that what u already said the chicken said?

Answer 18

that would be incorrect

Answer 19

LOL? Of course 0+0+0+0+0+0+0=0... 0 has no value..

Answer 20

@JOELMATHEWS1234 It's an indeterminate form.

Answer 21

infinity is just a concept, there's no arithmetic with it

Answer 22

So @hartnn , why would that be incorrect , is there some rule with indeterminates

Answer 23

You can do arithmetic with infinity.

Answer 24

Only with multiplying it to 0.

Answer 25

or dividing it against itself to equal 1.

Answer 26

@wio & @hartnn , according to tan graph , 1/0 = infinity , so is 0 * infinity = 1 ???

Answer 27

Just because \(0/0\) is an indeterminate form doesn't mean you can't do arithmetic with integers.

Answer 28

@ⒶArchie☁✪

Answer 29

So you object to: \[
\underbrace{0+0+\dots+0}_\infty=\infty \times 0
\]@hartnn

Answer 30

@hartnn

Answer 31

what i was saying is just , infinity is alias for 'very large number'
so, in that sense, when we add a very large numbers of 0, (we should get answer = 0)
thats left side
on right side, the notion is very large number times 0, which is also = 0
so in that sense i would agree to that

Answer 32

\[
\sum_{i=0}^n\frac 1n=1
\]\(\lim_{n\to\infty}\) Gives: \[
\underbrace{\frac 1\infty+\frac 1\infty+\dots+\frac 1\infty}_{\infty}=1
\]

Answer 33

Do you think it is impossible to add an infinite number of \(0\)s and eventually get something \(\neq 0\)?

Answer 34

@wio, I think that will give us infinity, not 1.

Answer 35

Nope, it gives \(n\times \frac 1n=1\)

Answer 36

thats sum is not 1

Answer 37

Make it \(i=1\)

Answer 38

If you have \(i=0\) it is \(1+\frac 1n\) but I mean to say \(i=1\).

Answer 39

still it has to be infinity, its diverging

Answer 40

Can you prove it is diverging?

Answer 41

Are you thinking of :\[
\sum_{i=1}^n\frac 1i
\]

Answer 42

oops, yes

Answer 43

Okay so the chicken is correct that in this case \(\lim_{n\to\infty}n\times 0 = 0\)

Answer 44

yes, the limiting value is indeed 0

Answer 45

But is it true that \[
\sum_{i=1}^nc=n\times c
\]Even when \(n\to \infty\)?

Answer 46

@Yttrium , the sum is 1 , you're adding 1 infinite times, and then dividing by infinite , so it is in fact 1

Answer 47

when n->infinity, it'll be c times infinity= infinity

Answer 48

Except when \(c\leq0\)? Are you sure though?

Answer 49

yeah, for negative c, it'll be negative infinity

Answer 50

Hmm, I'm wondering how one would prove it.

Answer 51

just plug in n= infinity :3

Answer 52

I think the clue lies in what is the definition of limit. The chicken though that since summation of any numbers of 0 is 0, the sum as n tends to infinity must be 0, right?

Answer 53

If you prove \[
\sum_{i=1}^nc=c\times n
\]It becomes trivial.

Answer 54

I do not agree with the chicken because the summation is 0 regardless of the number of zeros added. Limiting values are meant to be like asymptotic values that are approached as the number of zeroes increases and not meant to be a constant value obtained regardless of the value for n.

Answer 55

I don't think the limit is the tricky part.

Answer 56

yes - hold on... having connection problems atm... let me fix it. x_x

Answer 57

The chicken is right, but there is something in her proof that is wrong.

Answer 58

You don't see what's wrong with her proof?
Think of circular reasoning.
Also what you gave above is a generalization. The proof used for when c= 0 is the same you will use for a general c. You would use proof by induction. Apparently, you didn't seem to notice the mistake is the reasoning.

Answer 59

It's circular reasoning because the chicken used what she was trying to prove to prove what she needed to prove.

Answer 60

No the result wasn't used to prove itself.

Answer 61

\(p(k)\implies p(k+1)\) is part of induction.

Answer 62

the one underlined is what we are trying to prove. We cannot assume that it's equal to 0 which is what is assumed here. Okay. That's basically the logic to it.

Answer 63

You assume \(p(k)\) is true and use it to show \(p(k+1)\) is true.

Answer 64

Also that's the last step of the induction process. She skipped the middle step.
In that case, her proof would be correct.

Answer 65

What is the middle step?

Answer 66

The middle step is to assume that the general statement is true.

Answer 67

General statement?

There is only the base case and inductive hypothesis. There isn't any middle step.

Answer 68

2 and 3 are redundant but okay.

Answer 69

Struggling with latex?

Answer 70

The middle step is to assume that the general step is to assume that the general statement is true. The last step is to show that if you assume 2) is true. Then it's also true for the next term.
General statement is what we are trying to prove.
Which is on the board, i'm not sure if it's okay what the chicken say maybe it's assumed that we should know that the general term is assumed to be true.

Answer 71

lol, I'm using a different software from chrome web store its: Daum equation editor

Answer 72

Anyways, I'm not sure there is anymore to really say about this.

Answer 73

I made a new question, look at it!