jmc

We have that $$1+5x+9x^2+13x^3+\cdots =85.$$Multiplying both sides by $x,$ we get $$x+5x^2+9x^3+13x^4+\cdots =85x.$$Subtracting these equations, we get $$1+4x+4x^2+4x^3+4x^4+\cdots =85-85x.$$Then $$1+\frac{4x}{1-x}=85-85x.$$Multiplying both sides by $1-x,$ we get $$1-x+4x=(85-85x)(1-x).$$This simplifies to $85x^2-173x+84=0,$ which factors as $(5x-4)(17x-21)=0.$ Hence, $x=\frac{4}{5}$ or $x=\frac{21}{17}.$

In order for the series $1+5x+9x^2+13x^3+\cdots$ to converge, the value of $x$ must lie strictly between $-1$ and 1. Therefore, $x=\boxed{\frac{4}{5}}.$