<s>[INST] Proof that $P(x)=x-\frac{1}3 x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has radius of convergence $1$ 
Proof that
  $P(x)=x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has
  radius of convergence $1$

First of all, I need to convert this to a series:
$$\sum_{k=1}^\infty \frac{x^{2k-1}(-1)^k}{1-2k}$$
(I hope this is correct so far?)
On this series, you use ratio test, count the limit and then take the reciprocal of the limit which will hopefully equal 1:
$$\lim_{k\rightarrow\infty}\left (\frac{x^{2(k+1)-1}(-1)^{k+1}}{1-2(k+1)}: \frac{x^{2k-1}(-1)^k}{1-2k}\right ) = \lim_{k\rightarrow\infty}\frac{\left(x^{2k+1}(-1)^{k+1}\right)(1-2k)}{(-2k-1) (x^{2k-1}(-1)^k)}$$
$$=\lim_{k\rightarrow\infty}\left(\frac{x^{2k+1}}{x^{2k-1}}\right ) \cdot \left(\frac{(-1)^{k+1}}{(-1)^k}\right)\cdot\left(\frac{1-2k}{-2k-1}\right)$$
$$=\lim_{k\rightarrow\infty}\left (x^{2k+1-2k+1} \right )\cdot\left ((-1)^{k+1-k} \right )\cdot\left (\frac{k(\frac{1}{k}-2)}{k(-2-\frac{1}{k})} \right )$$
$$=\lim_{k\rightarrow\infty}x^2\cdot((-1)^k)\cdot\left (\frac{\frac{1}{k}-2}{-2-\frac{1}{k}} \right )=-x^2\cdot\left(\frac{0-2}{-2-0} \right) = -x^2$$
$$R=-\frac{1}{x^2}$$
After all, I don't get 1 as result... :(
Where is my mistake? I cannot imagine it's as complicated and long as I did?
Or maybe, at the end, I ignore the $x^2$? Then result would be 1!?
 [/INST] Taking the absolute value of your algebra gives $x^2$.  Since the ratio test looks for when this is $<1$, we ask ourselves when is $|x^2|<1$.  The answer is when $-1<x<1$, and if $|x|>1$ it does not converge.  This suffices to find the radius; it is $1$.  
If you want the interval of convergence as well, then you need to test when $|x|=1$.  In particular, just test $x=1$.  Then your sum becomes $$1-\frac13+\frac15-\frac17+\ldots$$
which must converge by the alternating series test.  Therefore, the interval of convergence is $-1 \leq x \leq 1$.  
Note, the radius is the same
</s>