Complex Zeroes

Let $2n=a$

Now, let's split this into cases:

Case 1: $a=\frac{p}{q}$ where $q$ is an odd number and $\gcd{(p,q)}=1$ , but $a$ still approaches infinity

When $|x|<1$ , $\lim _{ a\rightarrow \infty }{ \frac { x^{ a }-1 }{ x^{ a }+1 } } =\frac { 0-1 }{ 0+1 } =-1$

When $|x|>1$ , $\lim _{ a\rightarrow \infty }{ \frac { x^{ a }-1 }{ x^{ a }+1 } } =\frac { 1 }{ 1 } =1$

Therefore, points on the graph where $x=1$ and $x=-1$ are not continuous.

Case 2: $a=\frac{p}{q}$ where $q$ is an even number and $\gcd{(p,q)}=1$ ,but $a$ still approaches infinity

When $0 \le x <1$ , $\lim _{ a\rightarrow \infty }{ \frac { x^{ a }-1 }{ x^{ a }+1 } } =\frac { 0-1 }{ 0+1 } =-1$

When $x<0$ it is complex but as $a$ approaches infinity, $\Im m{\frac{x^{a}-1}{x^{a}+1}}=0$ . And it is easily derivable that again points on the graph where $x=1$ and $x=-1$ are not continuous.

Therefore, $1+-1=\boxed{0}$

Since no one's done it yet, I thought I'd post the graph. Graph

it's just find the sum of solution of the equation $x^{2n}+1=0$

even if we not take complex roots into account but since all the coefficient is real number so all complex root are in conjugated form.

Finally we have sum of answer $=\frac{-a_{n-1}}{a_{n}}=0$

-1 and 1 are the points of discontinuity

-1+1=0 Ans.