# I Proof that 0 + 0 +...+ 0 +... = 0

1. Nov 22, 2017

### jbriggs444

Note that there is a minor loss of rigor in the step where multiplication of the infinite sum by zero is shown as multiplication of the summand by zero. It turns out that multiplication by a fixed constant does distribute over [convergent] infinite sums. But that is something that would need to be proved rather than taken for granted.

2. Nov 22, 2017

### Staff: Mentor

Proving the first one is more effort than proving the last one.
And you also have to prove that you can commute the product with the series.
You are using a lot of advanced methods - methods that require knowledge of the last result already - to prove something very elementary.

3. Nov 22, 2017

### .Scott

I wasn't trying to create a Bertrand Russell type proof - only one that would build on what the OP would take as given.

4. Nov 26, 2017

### WWGD

A monotone Cauchy sequence ( assuming you have a countably-infinite sum) converges to its lub/glb?

5. Nov 26, 2017

Staff Emeritus
One of the issues here is "what is rigorous"? It's not a (no pun intended), 0/1 thing. (Okay, so maybe the pun was a little intended) There are degrees of rigor.

In most cases, what we have proven is that if this series converges to anything at all, it converges to 0. So there needs to be one more piece that shows it converges. Unfortunately, many convergence tests are stymied by the fact that all the terms are zero - for example, any test involving a ratio. Probably the way to prove this is to show that 1 + 1/2 + 1/4 + 1/8... converges to 2 and 0 + 0 + 0 + ... is strictly non-negative and uniformly smaller.

6. Nov 26, 2017

### WWGD

But you have a result for Real numbers that a Cauchy sequence converges. A Cauchy sequence is one where the difference between terns beyond a fixed one becomes "As small as you want", and it converges to the greatest lower bound.

7. Nov 27, 2017

### PeroK

Let $s_k = \sum_{n=1}^{k} 0$.

By definition: $\sum_{n=1}^{\infty} 0 = \lim_{k \rightarrow \infty} s_k$

(Note that $\forall k: s_k = 0$)

Let $\epsilon > 0$. Take $N =1$:

$k > N \Rightarrow |s_k - 0| = |0 - 0| = 0 < \epsilon$

Hence $\lim_{k \rightarrow \infty} s_k = 0$

QED

8. Nov 27, 2017

### lavinia

Any constant sequence in a metric space converges. There is nothing special about a sequence of finite sums of zeros. Also, one does not need the usual metric on the real numbers. Any metric will work.

The metric is used to interpret an infinite summation as a limit of a sequence. Without the metric one might still ask whether one can make sense of it.

Last edited: Nov 28, 2017
9. Nov 28, 2017

### Svein

Well, the standard way is to go by limits.

Let $S_{N}=\sum_{n=0}^{N}$. Then $(\forall N>0)(S_{N}=0)$. Thus $\sum_{n=0}^{\infty}=\lim_{N\rightarrow \infty}S_{N}=\lim_{N\rightarrow \infty}0=0$.

10. Dec 3, 2017

### Ssnow

If we correspond $0 \mapsto \emptyset$ and the binary operation $+\mapsto \cup$ then by the algebra of sets:

$\emptyset \cup \cdots \cup \emptyset \,=\, \emptyset$

that is

$0+ \cdots + 0\,=\, 0$.
Ssnow

11. Dec 4, 2017

### Staff: Mentor

With the same logic $1+ \cdots + 1\,=\, 1$ because $\{1\} \cup \cdots \cup \{1\} = \{1\}$?
That approach brings more problems than it is supposed to solve, especially as it is trivial to show the limit directly with the definitions (as done in posts 27 and 29).

12. Dec 5, 2017

### Ssnow

No beacuse defining the successor function for sets $S(A)=A\cup\{A\}$ you have that $1=S(\emptyset)=\emptyset \cup \{\emptyset\}=\{\emptyset,\{\emptyset\}\}$ so $1+1 \not = 1$ but $\{\{\emptyset,\{\emptyset\}\},\{\{\emptyset,\{\emptyset\}\}\}\}=2$.
Ssnow

13. Dec 5, 2017

### Staff: Mentor

That is different from what you suggested before, and it is still wrong. If you identify addition with the union of sets then x+x=x for all x because $s \cup s=s\quad \forall s$.

If you identify the addition with $f(s,t)=\{s,\{t\}\}$ then 1+1 works but nothing else works any more.

14. Dec 5, 2017

### Ssnow

Yes, I was thinking of "classes" rather than "sets" ...