Suppose that for every $n ∈ N$, $a_n > 0$ and set $b_n = \frac{a_{n+1}}{ a_n}$ and $c_n = (a_n)^{1/ n }$. Prove that if $b_n → L$ then $c_n → L$. Is the converse true.

If $a_n \to \infty$ then $b_n \to 1$ and similarly $c_n \to 1$

So it's easy to see this is the case. Seems like the squeeze theorem is the trick here. But I know $\neg(c_n \le b_n), \forall a_n$

but maybe: $ ??\le (a_n)^{1/n} \le (a_n)^{1/n} \cdot \frac{a_n+1}{a_n} $

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    $\begingroup$ Did you mean $b_n=\frac{a_{n+1}}{a_n}$? $\endgroup$ – Ángel Mario Gallegos Oct 6 '15 at 2:39
  • $\begingroup$ oh yes I do, looked at it wrong $\endgroup$ – oliverjones Oct 6 '15 at 2:40
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    $\begingroup$ assuming Mario G's correction, its not clear to me that $a_n \to \infty \implies b_n \to 1$ $\endgroup$ – user237392 Oct 6 '15 at 2:44
  • $\begingroup$ @bey I deleted that part after editing. $\endgroup$ – oliverjones Oct 6 '15 at 2:47

We give a semi-formal argument, for positive $L$. A slightly simpler argument works for $L=0$.

Let $\epsilon$ be very small positive, much smaller than $L$. There is an $N$ such that if $n\ge N$ then $L-\epsilon/2\lt \frac{a_{n+1}}{a_n}\lt L+\epsilon/2$.

It follows that for any $m$ we have $$a_N(L-\epsilon/2)^{m}\lt a_{N+m}\lt a_N(L+\epsilon/2)^m.$$ Take the $(N+m)$-th root. If $m$ is large enough, $a_N^{1/(N+m)}(L-\epsilon/2)^{m/(N+m)}$ is larger than $L-\epsilon$, and $a_N^{1/(N+m)}(L+\epsilon/2)^{m/(N+m)}$ is smaller than $L+\epsilon$.

So given $\epsilon\gt 0$, for large enough $k$ we have that $a_k^{1/k}$ is within $\epsilon$ of $L$.

Remark: The converse does not hold. Let $a_n=1$ if $n$ is odd, and let $a_n=2$ if $n$ is even.

  • $\begingroup$ This is a good approach that avoids the appeal to Stolz-Ceasaro, which the OP likely has not seen yet. +1 $\endgroup$ – Mark Viola Oct 6 '15 at 3:18

Let $x_n=\log b_n\to \log L$, then $$\log c_n=\frac{C+x_1+\cdots+x_n}n$$ for some constant C. We have a direct case of Cesaro mean.

  • $\begingroup$ This is a very efficient approach. So, +1. As a suggestion for improvement, you might consider expanding the explanation. That is, we can show that $\log c_n =\cdots $ from the fact that $a_2=b_1a_1$, $a_3=b_2a_2=b_2b_1a_1$ ... $a_n=a_1b_1b_2\cdots b_{n-1}$. Then, continue ... Make sense? $\endgroup$ – Mark Viola Oct 6 '15 at 3:16

Hint: Use the classical result whose proofs have been asked at least one time here: If $(a_{n})$ is a sequence of $a_{n} > 0$, then $$ \liminf \frac{a_{n+1}}{a_{n}} \leq \liminf_{n \to \infty}a_{n}^{1/n} \leq \limsup a_{n}^{1/n} \leq \limsup \frac{a_{n+1}}{a_{n}}. $$ From this it follows immediately that if $\lim_{n \to \infty}a_{n+1}/a_{n}$ exists and $=l$ for some $l \in \mathbb{R}$, then $\lim_{n \to \infty}a_{n}^{1/n}$ exists and $=l$.

  • $\begingroup$ how do we obtain the first inequality? $\endgroup$ – Nighty Oct 21 '15 at 14:43

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