Evaluate $1-x+x^2-x^3+\cdots$

Question

In some problem, I have to use the expression $$\sum^\infty_{k=0}(-1)^kx^k=1-x+x^2-x^3+\cdots$$ I know about Taylor series, but I'm not sure how to find the equivalent to this. It's similar to the $log(1+x)$ series. Any help will be appreciated.

Answer 1

For $|y| < 1$, $$ 1 + y + y^2 + \dots + y^n = \frac{1-y^{n+1}}{1-y} $$

so taking the limit $ n \to \infty$ we get

$$ \sum_{k=0}^\infty y^k = \frac{1}{1-y}. $$

Now take $y = -x$ and you get what you want.

Have I answered your question?

Answer 2

Since $(-1)^kx^k=(-x)^k$, you have a geometric series, whose sum is $\dfrac1{1+x}$, when $|x|<1$.

Answer 3

You’re exactly right.

The Taylor series for $\log(1+x)$ is $$\log(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$ so, differentiating both sides, you get $$\frac{1}{1+x} = 1-x+x^2-x^3+\cdots$$