$$
1+S(x)=x\cdot S(x)
$$
Let's we have the equation above
I noticed there are two solutions of that equation in series :

---

Solution 1:
$$
S\_1(x)=-\sum\_{n=0}^\infty x^n
$$
Showing that
$$
1+ (-1-x-x^2-x^3-....)=x(-1-x-x^2-x^3-....)
$$

---

Solution 2:
$$
S\_2(x)=\sum\_{n=1}^\infty x^{-n}
$$

Showing that
$$
1+ (x^{-1}+x^{-2}+x^{-3}+....)=x(x^{-1}+x^{-2}+x^{-3}+...)
$$

---

If $$
S\_1(x)=S\_2(x)
$$

We get 
$$
\sum\_{n=-\infty}^\infty x^n=0
$$

Second Method to Show:

$$
\sum\_{n=0}^\infty x^n=\frac{1}{1-x}
$$

$$
\sum\_{n=0}^\infty x^{-n}=\frac{1}{1-x^{-1}}
$$

$$
\sum\_{n=0}^\infty x^{-n}=\frac{x}{x-1}
$$

$$
-1+\sum\_{n=0}^\infty x^n+\sum\_{n=0}^\infty x^{-n}=\sum\_{n=-\infty}^\infty x^n=-1+\frac{1}{1-x}+\frac{x}{x-1}=-1+\frac{1}{1-x}-\frac{x}{1-x}
$$

$$
-1+\sum\_{n=0}^\infty x^n+\sum\_{n=0}^\infty x^{-n}=\sum\_{n=-\infty}^\infty x^n=-1+\frac{1-x}{1-x}=0
$$

$$
\sum\_{n=-\infty}^\infty x^n=0
$$

But If we test with a number put in $x$ 
 $$
\sum\_{n=-\infty}^\infty x^n 
$$
goes to $\infty$ not $0$

Where is the problem in my methods to show the sum is $0$?

Note: I know that the sum is divergence but I asked that question for some people that wonder why. I need good mathematical answer to show where the problem is.

Can it be similar to the example of [Ramanujan’s mysterious expression](https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF)?
 $$
(1+2+3+4+5+6+7+.....)=-\frac{1}{12}
$$.

Please see also [the article](https://plus.maths.org/content/infinity-or-just-112) that It mentions that the divergence series that is like above can get result as shown in [wiki page](https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF) .

$$
\sum\_{n=1}^\infty \frac{1}{n^x}=\zeta(x) 
$$

and according to article 
 $$
\zeta(-1) = -\frac{1}{12}
$$

Can it be same kind of thing in my result?

Please explain Ramanujan’s mysterious expression and my strange results.

Thanks