1
$\begingroup$

Using the ratio test, I keep getting a radius of convergence of $\left| z \right| < 2$, when the answer is 4. What leads to the answer of 4?

  • 0
    I assume you meant $n\to\infty$ rather than $i\to\infty$. – 2017-01-10

1 Answers 1

4

\begin{eqnarray} \lim_{n\to\infty}\left\lvert\dfrac{a_{n+1}}{a_n}\right\rvert&=&\lim_{n\to\infty}\left[\dfrac{(n+1)!}{n!}\right]^2\cdot\dfrac{(2n)!}{(2n+2)!}\vert z\vert\\ &=\lim_{n\to\infty}&\dfrac{(n+1)^2}{(2n+1)(2n+2)}\vert z\vert \end{eqnarray}

so clearly the limit goes to $\left\lvert\dfrac{z}{4}\right\rvert<1$.

So $\vert z\vert<4$

  • 1
    Ah thanks for that. I didn't distribute the 2 in (2(n+1))! in the denominator which explains my error. Thanks! – 2017-01-10