# Determining A Finite Value of Infinity

1. Nov 19, 2014

### willr12

okay.....if you accept that the sequence
1+2+3+4.....=-1/12,
I think I have determined a finite value of infinity.
To find the value of the sums of all natural numbers up to a number, you can use the equation
((x^2)+x)/2.
An example would be 4.
4+3+2+1=10.
((4^2)+4)/2 also equals 10.
following this logic,
((x^2)+x)/2=-1/12
is true for the above sequence. this can be rearranged to
(x^2)+x+1/6
This is the resulting quadratic equation. Using the quadratic formula, one obtains the x intercepts as
(-3+-(sqrt3))/6.
and since x is infinity in this situation (since the highest value is infinity), the x intercepts are the values of infinity in the equation. Therefore, by assigning a value of -1/12 to riemann zeta(-1), you also assign finite values to infinity, approximately
-0.211324....
and
-0.788675....
If there is any faulty reasoning, please remember I'm 15 and I most likely have no clue what I'm talking about.

2. Nov 19, 2014

### phinds

Since this is a contraction in terms, I stopped reading right there.

3. Nov 19, 2014

### jbriggs444

http://en.wikipedia.org/wiki/Divergent_series

There is no sum of the integers from one to infinity. The series is divergent. The assignment of the value -1/12 as the "sum" of this series does not correspond to anything which you should regard as being an actual sum.

4. Nov 19, 2014

### willr12

Most people would stop reading when someone says that 1+2+3.....=-1/12, that's counterintuitive but generally accepted.....so thanks for the ignorance

5. Nov 19, 2014

### willr12

Makes sense

6. Nov 19, 2014

### pwsnafu

No. The series 1+2+3+... is divergent. So it is infinite.
It's Zeta regularization has value -1/12.
That is not the same thing as claiming that the series is "equal" to -1/12

I fail to see how. Just because a statement is true for finite values of x does not mean it is valid for infinite values of x. You need to prove that "((x^2)+x)/2=-1/12" is a valid statement for some x. You have not.

This is not possible because (1) infinity is not a real number, (2) infinite values cannot be manipulated in the same way as finite values.

Unfortunately, everything.

7. Nov 19, 2014

### willr12

Good to know. Thanks for the reply

8. Nov 19, 2014

### phinds

No, as has been pointed out, it is NOT generally accepted.

9. Nov 19, 2014

### willr12

It is generally accepted as a representation of Riemann zeta (-1)

10. Nov 20, 2014

### pwsnafu

No you have this backwards.
$\zeta(-1)$ is a possible regularization of $\sum_{n=1}^{\infty} n$.
$\sum_{i=1}^{\infty} n$ is not a representation of $\zeta(-1)$.
Regularization and representation are separate concepts.

11. Nov 20, 2014

### willr12

Y-value of the vertex in the resulting quadratic equation is -1/12. Any significance do you think?
good to know thanks

12. Nov 21, 2014

### Dragonfall

I don't.

13. Nov 21, 2014

### Staff: Mentor

This thread has run its course, so I'm closing it.