Convergent alternating series problem Find the constant $~c~$  such that                                                          $$4c-4c^2+4c^3-4c^4+...=1 $$
So I took $~4c~$ as a common multiple and got
$$4c(1-c+c^2-c^3+...)=1~. $$
And now:
$$\sum_{k =0}^{\infty}(-1)^k c^k=1/4c$$
But then i didnt know what to do to find c.
P.s. Sorry iam bad with commands

You want to have $$ c-c^2+c^3-...=1/4$$
$$\frac {c}{1+c}= 1/4$$
Solve for $c$ and you get $c=1/3$
