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This question already has an answer here:

We were told, in recitation class, about a test for sequences convergence (not series) Which goes as follows:

if $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}=L$ then $\lim_{n \rightarrow \infty} \sqrt [n] {a_n}=L$.

In a previous question I asked: This limit: $\lim_{n \rightarrow \infty} \sqrt [n] {nk \choose n}$.

I was told that this fact is not true. My question is, can anyone think of a counter example for it? Because, If yes, Then I would like to let my tutor know about it, but, I don't want to doubt him befor I am sure of it.

Thank you!

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marked as duplicate by Martin Sleziak, egreg, Stefan4024, Dan Rust, MathOverview Nov 10 '13 at 14:41

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    $\begingroup$ You (or rather your tutor) was correct. See Theorem 2.21 here for a proof. $\endgroup$ – Casteels Nov 10 '13 at 12:41
  • $\begingroup$ See also this question or this question. (This seems to be duplicate of the first one.) $\endgroup$ – Martin Sleziak Nov 10 '13 at 14:05
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I assume we're dealing with (strictly) positive $a_n$ here. Then the assertion is true. Let

$$L = \lim_{n\to\infty} \frac{a_{n+1}}{a_n}.$$

Given $\varepsilon > 0$, choose $N\in\mathbb{N}$ so that

$$M := \max \{0,L-\varepsilon\} < \frac{a_{n+1}}{a_n} < L+\varepsilon$$

for all $n \geqslant N$. Then we have, for $n > N$

$$a_N\cdot M^{n-N} < a_n < a_N\cdot (L+\varepsilon)^{n-N},$$

and taking $n$-th roots

$$\sqrt[n]{a_N} \cdot M^{1-N/n} < \sqrt[n]{a_n} < \sqrt[n]{a_N}(L+\varepsilon)^{1-N/n}.$$

The limit of the lower bound is $M$, and the limit of the upper bound is $L+\varepsilon$, so

$$M \leqslant \liminf \sqrt[n]{a_n} \leqslant \limsup \sqrt[n]{a_n} \leqslant L+\varepsilon.$$

Since $\varepsilon$ was arbitrary, we have indeed

$$\lim_{n\to\infty} \sqrt[n]{a_n} = L.$$

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  • $\begingroup$ And this proves something that the statement isn't clear about, namely that $(\root n\of {a_n})_{n\in \Bbb N}$ converges. $\endgroup$ – Git Gud Nov 10 '13 at 12:50
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Note that the root requires $a_n\geq0$.

Fix $\varepsilon>0$. Then, for $n$ big enough, $$L-\varepsilon<\frac {a_{n+1}}{a_n}<L+\varepsilon. $$ Then $$ a_{n+1}\leq (L+\varepsilon)a_n\leq\cdots (L+\varepsilon)^{n}a_1. $$ So $$ \sqrt[n]{a_n}\leq(L+\varepsilon)(a_1)^{1/n}. $$ Then $\limsup_n\sqrt[n]{a_n}\leq(L+\varepsilon)$. As $\varepsilon$ was arbitrary, we get $\limsup_n\sqrt[n]{a_n}\leq L$.

In a similar way, from $a_{n+1}\geq (L-\varepsilon)a_n$ we get $\liminf_n\sqrt[n]{a_n}\geq(L-\varepsilon)$, so $\liminf_n\sqrt[n]{a_n}\leq L$.

The inequalities $$ L\leq\liminf_n\sqrt[n]{a_n}\leq\limsup_n\sqrt[n]{a_n}\leq L$$ show that the limit exists and equals $L$.

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  • $\begingroup$ Why do you take $\lim \sup$ and $\lim \inf$? Once you have $$a_1^{1\over n} (L -\varepsilon) < a_{n+1}^{1\over n} <a_1^{1\over n} (L + \varepsilon)$$ would it be wrong to just take the limit on all sides to get $$ L - \varepsilon \le \lim_{n \to \infty} a_{n+1}^{1\over n} \le L + \varepsilon ?$$ $\endgroup$ – student Sep 20 '16 at 14:12
  • $\begingroup$ No, because you don't know that $\lim a_{n}^{1/n}$ exists; and $L-\varepsilon\ne L+\varepsilon$. $\endgroup$ – Martin Argerami Sep 20 '16 at 14:54
  • $\begingroup$ Oh, ok, so because we don't know if that limit exists but we do know that lim inf and lim sup always exist we use those instead. Thank you! $\endgroup$ – student Sep 20 '16 at 15:20
  • $\begingroup$ No, wait, I'm still not clear. When I solved this I first derived $$a_{n + 1}^{1\over n + 1} ((L+\varepsilon)^{N})^{1\over n +1}< a_{N}^{1\over n + 1} (L + \varepsilon)$$ and $$ a_{N}^{1\over n+1} (L - \varepsilon) < a_{n + 1}^{1\over n+1} ((L-\varepsilon)^N)^{1\over n+1}$$ and then took the limit. At the time of limit taking I did not know the limit existed but I thought if I am given a sequence $a_n$ such that $a_n < K$ then also $\lim a_n \le K$. Or does this only hold if we already know $\lim a_n$ exists? $\endgroup$ – student Sep 20 '16 at 15:23
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    $\begingroup$ You are welcome :) $\endgroup$ – Martin Argerami Sep 20 '16 at 19:51

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