This may have been asked before, however I was unable to find any duplicate. This comes from pg. 52 of "Mathematical Analysis: An Introduction" by Browder. Problem 14:

If $(a_n)$ is a sequence in $\mathbb R$ and $a_n > 0$ for every $n$. Then show: $$ \liminf\frac{a_{n+1}}{a_n} \le \liminf a_n^{1/n} \le \limsup a_n^{1/n} \le \limsup\frac{a_{n+1}}{a_n}$$

The middle inequality is clear. However I am having a hard time showing the ones on the left and right. (It seems like the approach should be similar for each). This is homework, so it'd be great if someone could give me a hint to get started on at least one of the inequalities.


  • 3
    $\begingroup$ It may sound somewhat unhelpful, but the best technique is usually to ditch the tricks (at least at first) and go one step at a time with the definitions for every term. After a while that you have become comfortable with the definitions, it gets a lot easier and a lot clearer how "tricks" and other techniques work. $\endgroup$
    – Asaf Karagila
    Oct 2, 2011 at 23:56
  • 1
    $\begingroup$ See also these two answers: math.stackexchange.com/questions/76743/limit-of-fraca-n1a-n/… and math.stackexchange.com/questions/28476/… $\endgroup$ Nov 1, 2011 at 9:53
  • 1
    $\begingroup$ This is also given as Problem 2.4.26 in the book Kaczor, Nowak: Problems in Mathematical Analysis. The problem is stated on p.46 and a solution is given on p.205. $\endgroup$ Jul 23, 2012 at 5:55
  • 1
    $\begingroup$ Also in Ross, Elementary Analysis 2ed p.79-80 with solution to third inequality. $\endgroup$
    – user203509
    May 1, 2016 at 19:18
  • $\begingroup$ Actualy you can take $\ln$ on each term and use Stolz-Cesaro $\endgroup$
    – Tony Ma
    May 16, 2018 at 11:45

1 Answer 1


A somewhat detailed hint for the right hand side inequality:

Suppose $$ r := \lim \sup \frac{a_{n+1}}{a_n}. $$ (If the above expression is $\infty$, then there is nothing to prove. So assume $0 \leq r < \infty$.) Fix any $\epsilon > 0$. This means that there exists $N$ such that for $n \geq N$, we have $$ \frac{a_{n+1}}{a_n} \leq r + \epsilon. $$ From this, can you deduce that for $n \geq N$, we have $$ \frac{a_n}{a_N} \leq (r+\epsilon)^{n-N}? $$ Rearranging a bit, $$ a_n \leq (r+\epsilon)^{n} \left( \frac{a_N}{(r+\epsilon)^N} \right), $$ so that $$ a_n^{1/n} \leq (r+\epsilon) \left( \frac{a_N}{(r+\epsilon)^N} \right)^{1/n}. $$ Can you take it from here?

You do not have to work so hard for the left hand side inequality: there is a simple way to obtain the left hand side inequality using the right hand side one in a black-box way. I will leave you to figure it out.

  • 5
    $\begingroup$ Do we deduce $\frac{a_n}{a_N}\le(r+\varepsilon)^{n-N}$ from $\frac{a_{N+1}}{a_N}\cdot\frac{a_{N+2}}{a_{N+1}}\cdots\frac{a_{N+k}}{a_{N+k-1}}\le (r+\varepsilon)^k$? $\endgroup$
    – PinkyWay
    Aug 29, 2020 at 7:50
  • $\begingroup$ @Invisible how did you get n-N power.. $\endgroup$ Apr 6, 2023 at 14:50
  • $\begingroup$ i got $\frac{a_{N+k}}{a_N}$ $\endgroup$ Apr 6, 2023 at 14:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .