# Find a closed form for $\Sigma_{k = 1}^\infty 3x^{3k -1}$

Find a closed form form and the largest set on which this formula is valid.

a) $\Sigma_{k = 1}^\infty 3x^{3k -1}$

Attempt: $\Sigma_{k = 1}^\infty 3x^{3k -1} = 3\Sigma_{k = 1}^\infty x^{3k} x^{-1} = \frac{3}{x} \Sigma_{k = 1}^\infty (x^3)^k = \frac{3}{x} \frac{1}{1 - x^3}$

However this is incorrect. I don't know why it's incorrect. But the book's answer is $f(x) = \frac{3x^2}{(1 - x^3)}$ Can someone please show how do they get this answer. Thank you

Step 1: factor out $(3/x)$.
Step 2: Substitute $y=x^3$.
Result: $$(3/x)\sum_{k\ge 1} y^k=(3/x)\frac{y}{1-y}=(3/x)\frac{x^3}{1-x^3}$$
The error in OP was in the last step. The sum of a geometric series is the first term divided by the ratio subtracted from 1. This series begins with $k=1$, so the first term is $x^3$.
Note that $$\sum_{k=1}^\infty (x^3)^k = \frac{x^3}{1-x^3}$$ and the difference with $k$ starting from $0$ $$\sum_{k=0}^\infty (x^3)^k = \frac{1}{1-x^3}$$