Why $\sum\limits_{n=0}^\infty (-1)^nx^{2n}$ converges pointwise?

Question

Consider the geometric series $\displaystyle\sum_{n=0}^\infty (-1)^nx^{2n}$. Does it converges pointwise in the interval $-1<x<1$?

Solution.

The series has ratio $-x^2,$ thus the $n^{th}$ partial sum is $S_n=\dfrac{1-(-x^2)^{n+1}}{1+x^2}$.

For each $x\in(-1,1)$ this converges to $1/(1+x^2),$ so the series does converges pointwise.

My question is:

Why does the series converge to $1/(1+x^2)$?

Why does $-\dfrac{(-x^2)^{n+1}}{1+x^2}$ tend to zero when $n\to\infty$?

Answer 1

Why does $\frac{(-x^2)^{n + 1}}{1 + x^2}$ tends toward zero when $n$ goes to infinity?

The answer is because $-1 < x < 1$. This means that $x$ can be written as a number $x = \pm\frac{1}{y}$, such that $y > 1$. This gives:

$$ (-1)^{n + 1}\frac{y^{-2n - 2}}{1 + \frac{1}{y^2}} = (-1)^{n + 1}y^2\frac{y^{-2n - 2}}{1 + y^2} = (-1)^{n + 1}\frac{y^{-2n}}{1 + y^2} $$

This obviously tends toward zero since you have a negative exponential with a base, $y$, greater than 1.

Answer 2

Since domain is restricted in range$-1<x<1$ therefore if you tend the limit in your geometric series to $\infty$, then the number $x$ must converge to a particular point which is probably $0$ as you are raising a very high power on a fractional number which converges the sum to $\frac{1}{1+x^2}$. Hope my answer is clear.