jmc

From the formula for an infinite geometric series, $$1-x+x^2-x^3+\cdots =\frac{1}{1+x}.$$Thus, we want to solve $$x=\frac{1}{1+x}.$$This simplifies to $x^2+x-1=0.$ By the quadratic formula, $$x=\frac{-1\pm \sqrt{5}}{2}.$$The infinite geometric series $$1-x+x^2-x^3+\cdots$$converges only for $|x|<1,$ so the only solution in $x$ is $\boxed{\frac{-1+\sqrt{5}}{2}}.$