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The following is a problem from Chapter 22 "Infinite Sequences" of Spivak's Calculus

  1. Suppose that $a_n>0$ for each $n$ and that $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}=l\tag{1}$$ Prove that

$$\lim\limits_{n\to\infty} \sqrt[n]{a_n}=l\tag{2}$$

My solution differed from the solution manual, so I wanted to check its correctness.

Here is my solution attempt

$$a_n(l-\epsilon)<a_{n+1}<a_n(l+\epsilon)$$

$$a_{n+1}(l-\epsilon)<a_{n+2}<a_{n+1}(l+\epsilon)$$

$$a_n(l-\epsilon)^2<a_{n+2}<a_n(l+\epsilon)^2$$

Thus

$$a_n(l-\epsilon)^m<a_{n+m}<a_n(l+\epsilon)^m$$

$$\sqrt[n+m]{a_n(l-\epsilon)^m}<\sqrt[n+m]{a_{n+m}}<\sqrt[n+m]{a_n(l+\epsilon)^{m}}$$

$$\lim\limits_{m\to\infty}\sqrt[n+m]{a_n}\lim\limits_{m\to\infty}\sqrt[n+m]{(l-\epsilon)^m}<\lim\limits_{m\to\infty}\sqrt[n+m]{a_{n+m}}<\lim\limits_{m\to\infty}\sqrt[n+m]{a_n}\lim\limits_{m\to\infty}\sqrt[n+m]{(l+\epsilon)^m}\tag{1}$$

Since $a_n>0$, then $\lim\limits_{m\to\infty}\sqrt[n+m]{a_n}=1$.

$l>0$ since $a_n>0$. Note that $l\neq 0$ since then $(1)$ would be false.

Thus

$$\lim\limits_{m\to\infty}\sqrt[n+m]{(l-\epsilon)^m}=l-\epsilon$$

$$\lim\limits_{m\to\infty}\sqrt[n+m]{(l+\epsilon)^m}=l+\epsilon$$

and

$$l-\epsilon<\lim\limits_{m\to\infty} \sqrt[n+m]{a_{n+m}}<1+\epsilon$$

for every $\epsilon>0$. Thus

$$\lim\limits_{m\to\infty} \sqrt[n+m]{a_{n+m}}=l$$

Is this correct?

xoux
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  • I just noticed that I will need to reformulate what I wrote. I need to use an argument that uses terms in the sequence above some value $N$. I can't use things like $a_1$. So hold off please until I make this change. – xoux Oct 21 '22 at 06:00
  • Don't take limits on both sides of an inequality, unless you have proved the exsitence of the limits. I mean, $A_n\leq B_n$ doesn't imply $\lim_{n\to\infty}A_n\leq \lim_{n\to\infty}B_n$, unless they both exist. Here, you don't know the existence of $\lim \sqrt[n]{a_n}$ in advance. – Feng Oct 21 '22 at 06:03
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    Does this answer your question? If $a_{n+1}/a_n\to\ell$ show $a_n^{1/n}\to\ell$ - found using an Approach0 search. Note the OP's attempt there appears to be very similar to yours, with that question's accepted answer completing the argument. – John Omielan Oct 21 '22 at 06:11
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    @evianpring FYI, the search also turned up several posts that used this result, such as in this answer which defined & proved it as a lemma, as well as just stating it in this other question's answer, and in the AoPS thread's Infinite limit - Message #4. – John Omielan Oct 21 '22 at 06:16
  • @Feng I have reformulated by proof. I think the step you write about is now $(1)$. Take, for example, $\lim\limits_{m\to\infty} \sqrt[n+m]{a_n(l-\epsilon)^m}=\lim\limits_{m\to\infty} \sqrt[n+m]{a_n}\lim\limits{m\to\infty} \sqrt[n+m]{(l-\epsilon)^m}$. Proving that these individual limits exist can be done relatively easily. Then it is fine, right? – xoux Oct 21 '22 at 06:24
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    I was talking about $(1)$. But my point is the middle term of $(1)$: $\lim\limits_{m\to\infty}\sqrt[n+m]{a_{n+m}}$. We don't know the existence of this limit in advance. – Feng Oct 21 '22 at 06:29
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    I know this doesn't directly adress your question, but a stronger result is in https://math.stackexchange.com/questions/69386/inequality-involving-limsup-and-liminf-liminfa-n1-a-n-le-liminf – Anne Bauval Oct 21 '22 at 06:49
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    Note also that taking log your exercise is equivalent to Cesàro's theorem. – Anne Bauval Oct 21 '22 at 06:56

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