If $\limsup \frac{a_{n+1}}{a_n}<1$ does $\sum a_n$ converge even if $\lim\frac{a_{n+1}}{a_n}$ does not exist? I was wondering why the Ratio test has the $\lim$ sign and the root test the $\limsup$ sign.


*

*Quotient test: $\lim|\frac{a_{n+1}}{a_n}|<1\Rightarrow \sum a_n$
converges.

*Ratio test: $\limsup \sqrt[n]{|a_n|}<1\Rightarrow \sum a_n$
converges.


If I pick
$$a_n=\begin{cases}2^{-\frac{n}{2}} &\mbox{n even}\\3^{-\frac{n+1}{2}} &\mbox{n odd} \end{cases},$$
then for $\frac{a_n+1}{a_n}$ we either have $(3/2)^{\frac{n+1}{2}}$ or $(2/3)^{\frac{n}{2}}\cdot 3$. It is not bounded therefore the limit does not exist, and $\limsup=\infty$ makes no difference.
The textbook says that this example shows that a Quotient test analogously to the Ratio test with $\limsup|\frac{a_{n+1}}{a_n}|$ instead of $\lim|\frac{a_{n+1}}{a_n}|$ is not true. 
What does this sentence mean why does it justify that if we have $\limsup<1$ we do not have necessarily $\sum a_n<\infty$. Because that is what I am getting out of it. 
Edit:
I have understood it now the Ratio test also says if $\lim \frac{a_{n+1}}{a_n}>1$ the series diverges. If I would change this with $\limsup$ then we would get a contradiction with the root test
 A: The ratio test can be expressed as follows.
Let $\ell=\liminf_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$ and let $L=\limsup_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$.  Then, the series $$\sum_{n=1}^\infty a_n\begin{cases}
\text{converges (absolutey)}&, L<1\\\\
\text{diverges }&, \ell>1\\\\
\text{diverges }&, \left|\frac{a_{n+1}}{a_n}\right|\ge1\,\text{for all large}\,n\\\\
\text{inconclusive}&,\text{otherwise}
\end{cases}$$

The root test is stronger than the ratio test since
$$\liminf_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\le \liminf_{n\to \infty}\sqrt[n]{|a_n|}\le \limsup_{n\to\infty}\sqrt[n]{|a_n|}\le \limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$
In the example in the OP, the ratio test in inconclusive since $L=\infty$ and $\ell=0$.  However, the root test reveals
$$\limsup_{n\to\infty}\sqrt[n]{|a_n|}=2^{-1/2}<1$$
and the series converges.
A: If $\limsup_n\left\lvert\frac{a_{n+1}}{a_n}\right\rvert<1$, then take some $c\in\left(\limsup_n\left\lvert\frac{a_{n+1}}{a_n}\right\rvert,1\right)$. Then $\limsup_n\left\lvert\frac{a_{n+1}}{a_n}\right\rvert<c$ and so, for some $N\in\mathbb N$, if $n\geqslant N$, then $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert<c$. But then $\lvert a_{N+1}\rvert<c\lvert a_N\rvert$, $\lvert a_{N+2}\rvert<c^2\lvert a_N\rvert$ and so on. So, yes, the series $\sum_{n=0}^\infty a_n$ converges absolutely. There is no difference between the ratio test and the quotient test as far as this is concerned.
