# To show limit of $\left|\frac{ a_{n+1}}{a_n}\right|$ is smaller than lim inf $\left|(a_n)^{1/n}\right|$

Suppose that $a_{n}$ is a sequence of real numbers such that $\lim\limits_{n\to\infty} {\frac{\left|a_{n+1}\right|}{\left|(a_n)\right|}}$ exists, then $$\lim\limits_{n\to\infty} {\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}} \leq \liminf\limits_{n\to\infty} \left|a_n\right|^{1/n}$$.

This is how I have started:

Suppose $\lim\limits_{n\to\infty} {\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}}=L, L \in R$

Given any $\epsilon \gt 0$, there exist $n_0$ such that for all $n \ge n_0$, $$L - \epsilon \lt {\frac{\left|a_{n+1}\right|}{\left|a_{n}\right|}} \lt L+\epsilon$$

and after this I am quite clueless how to continue. Any help or hint is much appreciated. Please help if you could. Thank you very much.

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It may be helpful to see the relationships between the ratio test and root test lim infs and lim sups written out all at once. For this, see e.g. Proposition 11.29 of math.uga.edu/~pete/2400full.pdf – Pete L. Clark Apr 10 '13 at 15:15
@PeteL.Clark, can you please comment on my proof below? – user70532 Apr 10 '13 at 15:16

First note that if $\lim\left|{a_{n+1} \over a_n}\right| = L$, then $\liminf\left|{a_{n+1} \over a_n}\right| = L$. Let $A:= \{t : t < \liminf\left|{a_{n+1} \over a_n}\right|\}$. Given any $t \in A$, if we let $\epsilon := L - t$, then there exists an $N$ such that for all $n \geq N$, $L - \epsilon < \left|{a_{n+1} \over a_n}\right| < L + \epsilon$. Or $t < \left|{a_{n+1} \over a_n}\right|$ which is equivalent to $$|a_n| t < |a_{n+1}| \tag{1}.$$

Next, we will prove by induction that given any natural number $k$, $|a_n| t^k < |a_{n+k}|$. For the base case, note that the limit inequality holds for all $n \geq N$. Thus, $|a_{n+1}|t < |a_{n+2}|$. Note that if we multiply (1) by $t$ we get $|a_n |t^2 < |a_{n+1}|t < |a_{n+2}|$ proving the base case.

Second, if we assume $|a_n| t^k < |a_{n+k}|$ is true, multiplying by $t$ we get $|a_n |t^{k+1} < |a_{n+k}|t$. Similarly, by the limit inequality, $|a_{n+k}|t < |a_{n+k+1}|$. Thus, the statement is true by induction.

Next, we can take the $(n+k)^{th}$ root of both sides: $|a_n t^k|^{1/(n+k)} < |a_{n+k}|^{1/(n+k)} \implies |a_n|^{1/k}t^{k/(k+n)} < |a_{n+k}|^{1/(n+k)}$.

On the left hand side of the inequality, $$\lim_{k \to \infty} |a_n|^{1/k}t^{k/(k+n)} = t.$$ Thus, the limit infimum also equals $t$. Therefore, since the inequality of sequences are preserved in the limits of the sequences, ( http://www.proofwiki.org/wiki/Inequality_of_Sequences_Preserved_in_Limit)

if we take the limit infimum of both sides we get that $$t \leq \liminf_{n \to \infty}|a_{n}|^{1/n}.$$ Since our choice of $t$ was arbitrary, this inequality must hold for all $t \in A$. Thus, $\liminf_{n \to \infty}|a_{n}|^{1/n}$ is an upper bound for the set $A$ such that $$\liminf_{n \to \infty} \left|{a_{n+1} \over a_n}\right| = \lim_{n \to \infty} \left|{a_{n+1} \over a_n}\right| \leq \liminf_{n \to \infty}|a_{n}|^{1/n}$$ by the property of the least upper bound.

P.S. The argument with the set $A$ can be reformulated using this theorem: $$\forall \epsilon > 0 (x - \epsilon < y) \implies x < y$$ I changed this to the equivalent formulation of sets and upper bounds to avoid managing multiple $\epsilon$.

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