TITLE: Do the sequences from the ratio and root tests converge to the same limit?
QUESTION [8 upvotes]: For example, if we have: 
$$\sum_{n=1}^{\infty}a_n$$
Where the ratio test is satisfied. That is $\exists L$ s.t.
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L < 1$$
Does this mean that for the same $L$ that the root test's terms converge as well? That is, is it true that:
$$\lim_{n \to \infty} \left| a_n^{1/n} \right| = L$$
This is for a homework problem, but the homework is not to prove the general case; I was just wondering if this was true or not.

REPLY [7 votes]: The answer to your question is yes: for any $L \in [0,\infty]$, if $\frac{a_{n+1}}{a_n} \rightarrow L$, then also $a_n^{\frac{1}{n}} \rightarrow L$. 
Here is a stronger result: 
Theorem: For any sequence $\{a_n\}_{n=1}^{\infty}$ of positive real numbers one has
$$\liminf_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n \rightarrow \infty} 
a_n^{\frac{1}{n}} \leq \limsup_{n \rightarrow \infty} a_n^{\frac{1}{n}} \leq 
\limsup_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$$.  
For a proof see e.g. $\S 5.3$ of these notes.
Note that the converse is not true: it is possible for the root test limit to exist but the ratio test limit not to.  In fact we can get this just by mildly rearranging the terms of a convergent geometric sequence,e.g. 
$\frac{1}{2}, \frac{1}{8}, \frac{1}{4}, \frac{1}{32}, \frac{1}{16} \ldots$
Here half of the successive ratios are $\frac{1}{4}$ and the other half are $2$, so the ratio test limit does not exist.  But the root test limit is still $\frac{1}{2}$.