What is the series of the function 3 / ( 1- x^4)

Question

I know that $f(x) = \frac{1}{ 1-x } = \sum_{n=1}^\infty x^n$. We can find that $g(x) = \frac{1}{ 1-x^4 } = \sum_{n=1}^\infty (x^4)^n = \sum_{n=1}^\infty x^{4n}$. Does the sum converge? what is the convergence radius?

Answer 1

The series $ 1+t+t^2+\cdots$ has radius of convergence $1$, so converges if $|t|\lt 1$ and diverges if $|t|\gt 1$. Thus our series $1+x^4+x^8+\cdots$ converges if $x^4\lt 1$, and diverges if $x^4\gt 1$. So we have convergence if $|x|\lt 1$, divergence if $|x|\gt 1$.

Remark: Your sums should start at $n=0$, not at $n=1$. And $g(x)=3+3x^4+3x^8+ 3x^{12}+\cdots=\sum_{n=0}^\infty (3)(x^{4n})$.