Question

Find a power series representation for the function. (Give your power series representation centered at x = 0.) f(x) = x/6x^2 + 1 f(x) = sigma^infinity_n = 0 (-1)^n x^2n+1 6^n Determine the interval of convergence. (Enter your answer using interval notation.)

Answer 1

Looks like your function is

$f(x)=\dfrac x{6x^2+1}$

Rewrite it as

$f(x)=\dfrac x{1-(-6x^2)}$

Recall that for $|x|$, we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

If we replace $x$ with $-6x^2$, we get

$f(x)=\displaystyle x\sum_{n=0}^\infty\frac(-6x^2)^n=\sum_{n=0}^\infty (-6)^n x^{2n+1}$

By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-6)^{n+1} x^{2(n+1)+1}}{(-6)^n x^{2n+1}}\right|=6|x^2|\lim_{n\to\infty}1=6|x|^2$

Solving for $x$ gives the interval of convergence,

$|x|^2$

We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.