What is the closed form of infinite sum $x+x^4+x^7+x^{10}+...$? I know that $1+x+x^2+x^3+...=\frac{1}{1-x}$, $1+x^2+x^4+...=\frac{1}{1-x^2}$. It seems to follow some rules of geometric series. But I'm confused how to get the title from here. Thanks

It's always nice, if we want to see if the sum is geometric, to do factoring to ensure that the first term of the sum is $1$, so factor out an $x$:
$$x + x^4 + x^7 + x^{10} + ... = x(1 + x^3 + x^6 + x^9 + ...)$$
The parenthetical expression is a geometric series in which the ratio is $x^3$. Then as a result, by the formula for an infinite geometric sum,
$$1 + x^3 + x^6 + x^9 + ... = \frac{1}{1-x^3}$$
Thus, bringing in the factored-out $x$,
$$x + x^4 + x^7 + x^{10} + ... = \frac{x}{1-x^3}$$
Note that the formula for a geometric sum only holds when the ratio is less than $1$ in magnitude. Thus we require $|x^3| < 1$. (This can be simplified to $|x| < 1$ as $|x^n|=|x|^n$ and then you just take the cube root.)
