The question is simple: determine the radius of convergence for the given series. I have done this for other problems using the following rule which our book described:
If the given series is $$ \sum_{k=0}^{\infty} c_{k}x^k $$
Than the radius of convergence can be found using the following limit:
$$ R = \lim_{x\to\infty} \frac{c_{k}}{c_{k+1}} $$
The series I struggle with is given by:
$$ \sum_{k=0}^{\infty} \binom{2k}{k} x^k$$
This supposed answer to this question is that $R = \frac{1}{4}$, but my solution find $R$ to be $+\infty$. Here is my solution:
I rewrote the binomial formula using substitution $ \binom{2k}{k} = \frac{(2k)!}{k!k!} $, and plugged this in as $c_{k}$:
$$ R = \lim_{x\to\infty} \frac{\frac{(2k)!}{k!k!}}{\frac{(2k + 1)!}{(k + 1)!(k + 1)!}} $$ $$ R = \lim_{x\to\infty} \frac{(2k)!(k + 1)!(k + 1)!}{(2k + 1)!k!k!} $$
Some term can be simplified to:
$$ R = \lim_{x\to\infty} \frac{(k + 1)(k + 1)}{2k + 1} $$$$ R = \lim_{x\to\infty} \frac{k^2 + 2k + 1)}{2k + 1} $$
I now take the limit of the highest order terms
$$ R = \lim_{x\to\infty} \frac{k^2}{2k} $$ $$ R = \lim_{x\to\infty} \frac{k}{2} $$$$ R = \frac{\infty}{2} $$$$ R = \infty $$
As I said the supposed answer is $\frac{1}{4}$ and I think it is just a calculation mistake, but I can't seem to find it.
Thanks in advance, Matthias
$\endgroup$ 41 Answer
$\begingroup$You miscalculated $c_{k+1}$: it is $\dfrac{(2k+2)!}{(k+1)!(k+1)!}$, so you obtain$$\frac{c_k}{c_{k+1}}= \frac{(2k)!}{(2k+2)!}\, \frac{(k+1)!\,(k+1)!}{k!\,k!}= \frac{(k+1)^2}{(2k+1)(2k+2)}\sim_\infty \frac{k^2}{(2k)^2}=\frac14.$$
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