I think I found a flaw in Riemann Zeta Function Regularization I think I may have found a flaw in how Zeta Regularization works.
As we all know, it's very famous for proving that $1+2+3+4+...=(-1/12).$ See here

(5 rows of equations at the end of this post)
•On the first row I simply re-define zeta function as a different sigma sum using well established properties of math.
•On the second row I state an identity. Evidence
•On the third row I show $\zeta(-3)$ in the form mentioned on row 1, then via substitution and the identity on row 2 I am able to create a new valid representation of $\zeta(-3)$.
•On row 4 I state that the sum of all real numbers being $-1/12$ as "proven" using Riemann Zeta Function Regularization (see here)
•On row 5 I substitute in the $-1/12$ (from row 4) into the other representation of $\zeta(-3)$ created on row 3. This gives us a final value of $1/144$
The only problem is, for the past 150 years $\zeta(-3) = 1/120$, but now all of a sudden I've made $\zeta(-3) = 1/144$ and as we all know, $120≠144$?
$$
\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \sum_{n=1}^\infty n^{-s}
$$
$$
\sum_{n=1}^x n^3 = \left( \sum_{n=1}^x n \right)^2
$$
$$
\zeta(-3) = \sum_{n=1}^\infty n^3 = \left( \sum_{n=1}^\infty n \right)^2
$$
$$
\sum_{n=1}^\infty n = -\frac{1}{12}
$$
$$
\frac{1}{120} = \zeta(-3) = \left( - \frac{1}{12} \right)^2 = \frac{1}{144}
$$
 A: 
The only problem is, for the past 150 years $\zeta(-3) = 1/120$, but now all of a sudden I've made $\zeta(-3) = 1/144$ and as we all know, $120≠144$?

I'm gonna tell you right now the same thing I've told another guy a year or two ago: If you can 
accept that $\infty=\dfrac1{120}$ despite the fact that $\infty\neq\dfrac1{120}$, then you can also accept that $\dfrac1{120}=\dfrac1{144}$ 
despite the fact that $\dfrac1{120}\neq\dfrac1{144}.~$ As boolean logic informs us, a falsehood can imply anything. 
After all, aren't $\dfrac1{120}$ and $\dfrac1{144}$ far closer to one another than either of them is to $\infty$ ? ;-$)$ If you 
have already accepted the bigger lie, what's to stop you from tolerating smaller untruths ? $($On a 
somewhat more serious note, in the holy war against divergent series, self-consistency is often 
the first to fall victim$)$.
A: When I read the above upvoted answer, I remember the hostility when I posted my first question related to this topic (applied on product series). It was discarded as total nonsense and the topic voted to be closed. Yet it was numerically beautiful. Eventually I figured it out myself.
When I first heard about assigning a finite value to divergent series I was intrigued. If something is so commonly accepted as this, the least you can do is think why it makes sense for other people who, most likely, are wiser than you. If you say it doesn't make sense you need at least to understand why it does make sense for someone else. Since this is a question, posting the assigned value, doesn't make sense without providing a supposed answer doesn't make sense to me. The question wasn't if assigning a finite value was correct or not, the question was why the squaring identity seems to give another value assumed the given value is correct.
So I'll give it my shot. I hope you find this useful.
A way to show the given identity is still true is to start from the beginning of expanding your function and take special care when you let your limits take over the action.
Now, $d$ is the period of expanding, for every $d>1$ you will get the same constant so $d=2$ is the most easy to calculate with. And $p$ will go to infinity.
$$\sum_{n=1}^{dp} f(n)\sum_{k=0}^{d-1} (e^{\frac{2i\pi k}{d}})^{n}=\sum^p_{n=1} d f(nd)$$
$$\sum_{n=1}^{\infty} f(n)\sum_{k=1}^{d-1} (e^{\frac{2i\pi k}{d}})^{n}=\sum^{\infty}_{n=1} (d f(nd)-f(n))$$
$$\sum_{n=1}^{\infty} n((-1)^n)=\sum^{\infty}_{n=1} 3n$$
Now you got an alternating form and it's Cesàro summable for example.
$$\sum_{n=1}^{\infty} n((-1)^n)=-1/4$$
$$-1/12=\sum^{\infty}_{n=1} n$$
Until this point everything is still fine:
$$\sum_{n=1}^{dp} f(n)\sum_{k=0}^{d-1} (e^{\frac{2i\pi k}{d}})^{n}=\sum^p_{n=1} d f(nd)$$
Now you assume incorrectly(!) you can make the following step if you still wish to square it.
$$\sum_{n=1}^{\infty} f(n)\sum_{k=1}^{d-1} (e^{\frac{2i\pi k}{d}})^{n}=\sum^{\infty}_{n=1} (d f(nd)-f(n))$$
Don't get me wrong the step is essential when you wish to get $p$ to infinity, but you can't square it any more after you take the limit to infinity.
The correct way is to take the square from here and you will see you will get the result.
$$(\sum_{n=1}^{dp} n*(1+(-1)^n))^2=(\sum^p_{n=1} 4n)^2$$
$$(\sum_{n=1}^{dp}n)^2+ (\sum_{n=1}^{dp}n*(-1)^n)^2+2(\sum_{n=1}^{dp}n*(-1)^n)(\sum_{n=1}^{dp}n)  =16\sum^p_{n=1} n^3$$
Since the $2(\sum_{n=1}^{dp}n*(-1)^n)(\sum_{n=1}^{dp}n)$ is irrelevant because of the order of the sum, it's growing yet don't got a constant nor dominant part. You can completely neglect it.
$$(\sum_{n=1}^{\infty}n)^2+ (\sum_{n=1}^{dp}n*(-1)^n)^2  =16\sum^{\infty}_{n=1} n^3$$
$$(\sum_{n=1}^{dp}n*(-1)^n)^2 =15\sum^{\infty}_{n=1} n^3$$
Take note of the alternating part that will become real since you square it, hence the 2 in 2/16.
$$2/16 =15\sum^{\infty}_{n=1} n^3$$
$$1/120 =\sum^{\infty}_{n=1} n^3$$
And to check it, we start with the function $f(n)=n^3$.
Now since we aren't going to square it, we can indeed start with:
$$\sum_{n=1}^{dp} f(n)\sum_{k=1}^{d-1} (e^{\frac{2i\pi k}{d}})^{n}=\sum^p_{n=1} (d f(nd)-f(n))$$
$$\sum_{n=1}^{\infty}n^3 (-1)^n =\sum^{\infty}_{n=1} 15n^3$$
Now you got an alternating form.
$$\sum_{n=1}^{\infty} n^3((-1)^n)=1/8$$
$$1/120 =\sum^{\infty}_{n=1} n^3$$
It's fascinating to see that this identity still works and while the intermediate results are different yet give the same answer. 
