Ratio test and the Root test Both the ratio test and the root test define a number (via a limit).
If both limits exist (and shows that the series is convergent), what (if any) is the relation between the 2 numbers ? are they equal ? 
What is the relation (if any) between them and the original series (other than the fact that they say the series is convergent) ?
 A: For a non-negative real series $(a_n)_{n \in \mathbb{N}}$, the tests give two (possibly undefined) numbers: let’s call them $L_\textit{root} := \lim_n (a_n)^{\frac{1}{n}}$, and $L_\textit{ratio} := \lim_n \frac{a_{n+1}}{a_n}$.
From Lemma 3 of these notes by Pete L. Clark, it follows that if $L_{\textit{ratio}}$ is defined, then $L_\textit{root}$ is also defined, and they are equal.
This is reasonably intuitive, with a bit of thought: suppose that for $n>N$, the ratio of consecutive terms $\frac{a_{n+1}}{a_n}$ is always close to $L$. Then (still for $n>N$), consider $a_n$ as produced by multiplying $a_N$ by all the later consecutive ratios; so it’s close to $L^{n-N} a_N$, and its $n$th root is close to $(L^{n-N} a_N)^{\frac{1}{n}} = L (\frac{a_N}{L^N})^\frac{1}{n}$.  The second factor here, being the $n$th root of a constant, goes to $1$ as $n$ grows; so for sufficiently large $n$, $(a_n)^\frac{1}{n}$ will be close to $L$.  (Exercise: make this argument precise — replace each “…close to…” by appropriate specific bounds.)
On the other hand, the converse doesn’t generally hold. $L_\textit{root}$ may be defined even if $L_\textit{ratio}$ is not.  For example, set $a_n = 2^n$ when $n$ is even, $a_n = 2^{n-1}$ when $n$ is odd.  Then the ratio of consecutive terms alternates between 1 and 4, so $L_\textit{ratio}$ is undefined; but the sequence is close enough to $2^n$ that the root converges, with $L_\textit{root} = 2$.
(Thanks to @David Mitra’s comment for the reference to the linked notes.)
A: For your first question:
If both limits exist, they must be equal to each other. In fact, for a sequence of positive terms $(a_n)$, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does
$\lim\limits_{n\rightarrow\infty}\root n \of {a_n}$ and moreover, in this case, the two limits are equal to each other. This follows from a more general fact contained in these notes of Pete L. Clark.

I'm not sure if the following answers your second question, but:
In general, there is no relationship between the value of the limit $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ and the value of the sum $\sum\limits_{n=1}^\infty a_n$.
Indeed, here is a silly example showing this:
Suppose $(a_n)$ is a sequence of positive terms and that $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}=r<1$. Then $\sum\limits_{n=1}^\infty a_n$ converges, say to $S\ne 0$.  Now let $a>0$ and  consider the sequence $(b_n)$ defined by $b_n=a\cdot a_n$. Here we have
$\lim\limits_{n\rightarrow\infty} {b_{n+1}\over b_n} =\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}= r$.  But,
$\sum\limits_{n=1}^\infty b_n=aS$.
So if, $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}=r<1$, the corresponding series could possibly converge to any given positive number.  The same remark holds for the limit in the Root test.
A: Both tests will yield the same property about the series:


*

*if $L<1$ the series is absolutely convergent

*if $L>1$ the series is divergent.

*if $L=1$ the series may be divergent, conditionally convergent, absolutely convergent


Edit: Note that if L=1 in the ratio test, the root test will also yield L=1. The converse is not true. (I had mistakingly asserted that it was earlier.)
As for relation... I'm not sure what you mean. The root test can be considered more comprehensive as it yields information whenever the ratio test is inconclusive. Applying the ratio test, however, can simpler in certain cases or perhaps necessary like when dealing with factorials.
