Expand dfrac14-x^4 in a power series in x, using known

Expand $\dfrac1{4-x^4}$ in a power series in $x$, using known expansions, and indicate when it is valid.

Step 1 · Factor

$$ \frac1{4-x^4} = \frac1{4\left(1-\frac{x^4}4\right)} = \frac14\cdot\frac1{1-\frac{x^4}4}. $$

Step 2 · Geometric series

With $t=x^4/4$: $$ \frac1{4-x^4} = \frac14\sum_{n=0}^{\infty}\left(\frac{x^4}4\right)^n = \sum_{n=0}^{\infty}\frac{x^{4n}}{4^{n+1}} = \frac14+\frac{x^4}{16}+\frac{x^8}{64}+\cdots $$

Step 3 · Region of convergence

$$ \left|\frac{x^4}4\right|<1 \iff |x|^4<4 \iff |x|<\sqrt[4]4=\sqrt2. $$

Summary Table

Additional Notes

• This is exactly Problem 247's family $1/(1+x^k)$ in disguise, up to rescaling: writing $4-x^4=4(1-(x/\sqrt2)^4)$ shows this is the $k=4$ case with $x$ rescaled by $\sqrt2$, matching the radius found here.
• The four singularities of $1/(4-x^4)$ are the fourth roots of $4$: $\pm\sqrt2$ and $\pm i\sqrt2$; the radius of convergence, $\sqrt2$, is exactly the distance from the origin to the nearest of these.
• Alternatively, factoring $4-x^4=(2-x^2)(2+x^2)$ and applying partial fractions gives the same series via two simpler geometric expansions in $x^2$ — a useful cross-check technique also used in Problem 260.