Root test is stronger than ratio test? I am a little bit confused regarding the meaning of the phrase :" Root test is stronger than ratio test", and was hoping you will be able to help me figure it out. As far as I can see here: https://www.maa.org/sites/default/files/0025570x33450.di021200.02p0190s.pdf
The limit from the ratio test is greater or equal the limit from the root test . So, my first question is-  is there any example of a series $\Sigma a_n$ such that the limit from the ratio test is exactly 1 (i.e.- inconclusive), but the limit from the root test is less than 1? (i.e.- convergence can be proved by using the root test but not by using the ratio test )
If not, then is it correct that this phrase is the meaning of "stronger" is when the limit from the ratio test does not exist? (as in the classic example of a rearranged geometric series)
Hope you will be able to help. 
THanks ! 
related posts: 
Show root test is stronger than ratio test
Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$
Do the sequences from the ratio and root tests converge to the same limit?

Consider the example of series
$$\sum 3^{-n-(-1)^n}$$
root test establishs the convergance but ratio test fails
onother example series with nth term
$a_n=2^{-n}$ if n is odd
$a_n=2^{-n+2}$ if n is even
for second series
when n is odd or even and tends to $\infty$
${a_n}^{\frac{1}{n}}=\frac{1}{2}$
Hence by cauchys root test the series converges
but the ratio test gives $\frac{a_n}{a_n+1}=\frac{1}{2}$ if n is odd and tends to $\infty$
$\frac{a_n}{a_n+1}=8$ when n is even and approachs $\infty$
Hence ratio test fails..
Sorry I dnt know mathjax that is why i was a bit late...
