Radius of convergence - ratio test for power series/real numbers

Having trouble with when to apply the ratio test for power series and when to apply the ratio test for real numbers.

For example, find radius of convergence of these....

$\sum_{n=0}^{\infty}(-1)^n (x+3)^n$

For this I would use ratio test for power series and get radius of convergence = 1, fine.

but for

$\sum_{n=0}^{\infty}n!(n+1)!(x+1)^{2n}$

Why do we need to use the ratio test for real numbers, and define for a fixed $x$? and why can't we use the ratio test for power series?

Is it because the $(x+1)^{2n}$ part is $2n$ and not $n$?

• $\sum_{n=0}^{\infty} n!(n+1)!(x+1)^{2n}$ is a power series in $(x+1)^2$. The ratio test for power series will work fine and give a radius of $0$. Latex is explained at [maa.org/latex/ltx-2.html]. May 15, 2013 at 12:42
• but its 2n...so all the odd powers of x are missing. I think that's maybe why. radius is 0 though...
– Tom
May 15, 2013 at 13:02
• when coefficients are zero, you cannot put them in the denominator as required in the ratio test. In some cases (as here) you can recognize it as a power series in another variable and use the ratio test on that one. May 15, 2013 at 14:07

Let $$\sum _{n=0} ^{\infty} n!(n+1)!(x+1)^{2n}$$ wants be convegent, so according to tests, we should have:
$$\lim_{n\to \infty}{\Big|\frac{(n + 1)! (n+2)! (x+1)^{(2n + 2)}}{n!(n+1)!(x+1)^{2n}}\Big|}<1$$ So,
$$\lim_{n\to \infty}{|(n + 1)(n + 2)x^2|}<1$$
Since for each $$x\ne0$$ we see $$\lim_{n\to \infty}{|(n+1)(n+2)x^2|}=\infty$$ so only for $$x=0$$ our series will be convergent and hence radius of convergence is equal to zero.