The improved ratio test says that given a series $\sum_{n=1}^{\infty}x_{n}$ with $x_{n}\geq0$ for all
$n\in\mathbb{N}$, if$$
\limsup_{n\rightarrow\infty}\sqrt[n]{x_{n}}<1,
$$
then the series converges, while if
$$
\limsup_{n\rightarrow\infty}\sqrt[n]{x_{n}}>1,
$$
then the series diverges to $\infty$.
On the other hand, the ratio test says that given a series $\sum_{n=1}^{\infty}x_{n}$ with $x_{n}>0$ for all
$n\in\mathbb{N}$, if$$
\limsup_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}<1,
$$
then the series converges. If
$$
\liminf_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}>1,
$$
then the series diverges to $\infty$.
Now given a sequence $\left\{ x_{n}\right\} $
of real numbers, with $x_{n}>0$ for all $n\in\mathbb{N}$. Then (see inequalities
$$
\liminf\limits_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}\leq\liminf
\limits_{n\rightarrow\infty}\sqrt[n]{x_{n}}\leq\limsup\limits_{n\rightarrow
\infty}\sqrt[n]{x_{n}}\leq\limsup\limits_{n\rightarrow\infty}\frac{x_{n+1}
}{x_{n}}.
$$
All the inequalities can be strict. It follows that if $\lim\limits_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}=\ell$ exists then there exists $\lim\limits_{n\rightarrow
\infty}\sqrt[n]{x_{n}}=\ell,$ so the two tests give the same result.
However in general the the ratio test is worse
than the root test.
Consider the series $\sum_{n=1}^{\infty}x_{n}$, where $$
x_{n}=\left\{
\begin{array}
[c]{ll}%
\frac{1}{2^{n}} & \text{if }n\text{ is even,}\\
\frac{1}{3^{n}} & \text{if }n\text{ is odd.}%
\end{array}
\right.
$$
Then $$
\liminf\limits_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}=0,\quad
\limsup\limits_{n\rightarrow\infty}\frac{x_{n+1}}{x_{n}}=\infty,
$$
and so the ratio test gives NO information. However,$$
\limsup\limits_{n\rightarrow\infty}\sqrt[n]{x_{n}}=\frac{1}{2},
$$
and so the series converges by the root test.
