Theorem: convergence for the ratio test implies convergence for the
root test. So whenever the ratio test works (i.e. tells you whether
the series converges), the root test also works and the limits
coincide.If $$\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}=L,$$ then
$$\lim_{n\to\infty}|a_n|^{1/n}=L.$$
Question: a series that cannot be evaluated through the root test because algebra won't allow it (usually involving a factorial), but can be evaluated through the ratio test will it theoretically have the same limit value through root test (if it were possible to be evaluated) as ratio test due to the theorem?
Example:
$$\displaystyle\sum\frac{n!}{n^n}$$
ratio test yields $ L = \frac{1}{e} < 1$
but if you tried the root test, you would be miserable. The limit is 1/e by ratio test implies root test?
(though you can just use Stirling's)
$$\frac{\sqrt{2\pi n} e^{-n} n^n}{n^n} \\
= \sqrt{2\pi} \cdot \frac{1}{e^n} \cdot \sqrt{n}$$The limit $1 \le \sqrt{n}^{1/n} \le n^{1/n} \to 1$ as $n \to +\infty$
allows us to recover the ratio $1/e$$$L = \lim\limits_{n\to+\infty} a_n^{1/n} = \lim\limits_{n\to+\infty} \frac1e \sqrt{n}^{1/n} = \frac1e$$
Best Answer
The ratio test can fail but the root test can succeed making the root test stronger than the ratio test.
$ \sum_{n=1}^{\infty} 2^{(-1)^n-n}$