This is less interesting of a result than may initially seem but I would still pat you on the back for successfully making a correct statement about divergent series. These are non-intuitive objects so even being able to re-state the obvious through an unusual line of a thinking helps at the very least develop your creativity.

Recall that:

$$ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}, |x|< 1 $$

The divergent series observation here is that we assert in general that:

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

Even when $|x| \ge 1$. All other techniques such as Cesaro, Abel, Borel, Ramanujan, Euler-Maclarin summation etc.. are designed to be consistent with THIS fundamental result. If you make a summation method and it is assigns a finite result that deviates from this then your summation method is NOT compatible with analytic continuation and therefore probably useless (and that is why you're technique was compatible with them).

From here we consider the series

$$ \sum_{n=0}^{\infty} e^{(2n+1)i\cos^{-1}(x)} = e^{i\cos^{-1}(x)} \sum_{n=0}^{\infty} \left(e^{2i\cos^{-1}(x)} \right)^n = \frac{ e^{i\cos^{-1}(x)} }{1-e^{2i\cos^{-1}(x)}}$$

We recall that $e^{i \cos^{-1}(x)} = x +i \sqrt{1 - x^2}$ from Euler's Formula and therefore this simplifies to

$$ \frac{ e^{i\cos^{-1}(x)} }{1-e^{2i\cos^{-1}(x)}} = \frac{i}{2\sqrt{1-x^2}}$$

With the help of Wolfram Alpha. Although showing that algebraically is a bit more difficult than meets the eye, despite being merely grade school algebra.

From here we can note:

$$ \frac{i}{2\sqrt{1-x^2}} = \frac{1}{(-i)2\sqrt{1-x^2}} \underbrace{=}_{\text{this requires a bit more justification}} \frac{1}{2\sqrt{(-1)(1 -x^2)}} = \frac{1}{2\sqrt{x^2-1}} $$

Which is the result you have intuitively found.

Once again, BECAUSE this result is just geometric series formula under the hood, it therefore MUST be compatible with all the famous summation techniques.

This is less interesting of a result than may initially seem but it's still commendable to find a correct statement about divergent series. These are non-intuitive objects so even being able to re-state the obvious through an unusual line of a thinking helps at the very least develop your creativity.

Recall that:

$$ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}, |x|< 1 $$

The divergent series observation here is that we assert in general that:

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

Even when $|x| \ge 1$. All other techniques such as Cesaro, Abel, Borel, Ramanujan, Euler-Maclarin summation etc.. are designed to be consistent with THIS fundamental result. If you make a summation method and it is assigns a finite result that deviates from this then your summation method is NOT compatible with analytic continuation and therefore probably useless (and that is why you're technique was compatible with them).

From here we consider the series

$$ \sum_{n=0}^{\infty} e^{(2n+1)i\cos^{-1}(x)} = e^{i\cos^{-1}(x)} \sum_{n=0}^{\infty} \left(e^{2i\cos^{-1}(x)} \right)^n = \frac{ e^{i\cos^{-1}(x)} }{1-e^{2i\cos^{-1}(x)}}$$

We recall that $e^{i \cos^{-1}(x)} = x +i \sqrt{1 - x^2}$ from Euler's Formula and therefore this simplifies to

$$ \frac{ e^{i\cos^{-1}(x)} }{1-e^{2i\cos^{-1}(x)}} = \frac{i}{2\sqrt{1-x^2}}$$

With the help of Wolfram Alpha. Although showing that algebraically is a bit more difficult than meets the eye, despite being merely grade school algebra.

From here we can note:

$$ \frac{i}{2\sqrt{1-x^2}} = \frac{1}{(-i)2\sqrt{1-x^2}} \underbrace{=}_{\text{this requires a bit more justification}} \frac{1}{2\sqrt{(-1)(1 -x^2)}} = \frac{1}{2\sqrt{x^2-1}} $$

Which is the result you have intuitively found.

Once again, BECAUSE this result is just geometric series formula under the hood, it therefore MUST be compatible with all the famous summation techniques.

This is less interesting of a result than may initially seem but I would still pat you on the back for successfully making a correct statement about divergent series. These are non-intuitive objects so even being able to re-state the obvious through an unusual line of a thinking helps at the very least develop your creativity.

Recall that:

$$ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}, |x|< 1 $$

The divergent series observation here is that we assert in general that:

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

Even when $|x| \ge 1$. All other techniques such as Cesaro, Abel, Borel, Ramanujan, Euler-Maclarin summation etc.. are designed to be consistent with THIS fundamental result. If you make a summation method and it is assigns a finite result that deviates from this then your summation method is NOT compatible with analytic continuation and therefore probably useless (and that is why you're technique was compatible with them).

From here we consider the series

$$ \sum_{n=0}^{\infty} e^{(2n+1)i\cos^{-1}(x)} = e^{i\cos^{-1}(x)} \sum_{n=0}^{\infty} \left(e^{2i\cos^{-1}(x)} \right)^n = \frac{ e^{i\cos^{-1}(x)} }{1-e^{2i\cos^{-1}(x)}}$$

We recall that $e^{i \cos^{-1}(x)} = x +i \sqrt{1 - x^2}$ from Euler's Formula and therefore this simplifies to

$$ \frac{ e^{i\cos^{-1}(x)} }{1-e^{2i\cos^{-1}(x)}} = \frac{i}{2\sqrt{1-x^2}}$$

With the help of Wolfram Alpha. Although showing that algebraically is a bit more difficult than meets the eye, despite being merely grade school algebra.

From here we can note:

$$ \frac{i}{2\sqrt{1-x^2}} = \frac{1}{(-i)2\sqrt{1-x^2}} = \frac{1}{2\sqrt{(-1)(1 -x^2)}} = \frac{1}{2\sqrt{x^2-1}} $$

Which is the result you have intuitively found.

Once again, BECAUSE this result is just geometric series formula under the hood, it therefore MUST be compatible with all the famous summation techniques.

This is less interesting of a result than may initially seem but I would still pat you on the back for successfully making a correct statement about divergent series. These are non-intuitive objects so even being able to re-state the obvious through an unusual line of a thinking helps at the very least develop your creativity.

Recall that:

$$ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}, |x|< 1 $$

The divergent series observation here is that we assert in general that:

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

Even when $|x| \ge 1$. All other techniques such as Cesaro, Abel, Borel, Ramanujan, Euler-Maclarin summation etc.. are designed to be consistent with THIS fundamental result. If you make a summation method and it is assigns a finite result that deviates from this then your summation method is NOT compatible with analytic continuation and therefore probably useless (and that is why you're technique was compatible with them).

From here we consider the series

$$ \sum_{n=0}^{\infty} e^{(2n+1)i\cos^{-1}(x)} = e^{i\cos^{-1}(x)} \sum_{n=0}^{\infty} \left(e^{2i\cos^{-1}(x)} \right)^n = \frac{ e^{i\cos^{-1}(x)} }{1-e^{2i\cos^{-1}(x)}}$$

We recall that $e^{i \cos^{-1}(x)} = x +i \sqrt{1 - x^2}$ from Euler's Formula and therefore this simplifies to

$$ \frac{ e^{i\cos^{-1}(x)} }{1-e^{2i\cos^{-1}(x)}} = \frac{i}{2\sqrt{1-x^2}}$$

With the help of Wolfram Alpha. Although showing that algebraically is a bit more difficult than meets the eye, despite being merely grade school algebra.

From here we can note:

$$ \frac{i}{2\sqrt{1-x^2}} = \frac{1}{(-i)2\sqrt{1-x^2}} \underbrace{=}_{\text{this requires a bit more justification}} \frac{1}{2\sqrt{(-1)(1 -x^2)}} = \frac{1}{2\sqrt{x^2-1}} $$

Which is the result you have intuitively found.

Once again, BECAUSE this result is just geometric series formula under the hood, it therefore MUST be compatible with all the famous summation techniques.