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Suppose we have a sequence $(a_n)$ such that $a_n\neq0$ and $$\lim\dfrac{a_{n+1}}{a_n}=\ell$$ Show that $$\lim a_n^{1/n}=\ell.$$

I have a proof sketch, but it needs an extra assumption:
By the definition of convergence $\forall\varepsilon>0\exists N\forall n>N,\ \ell-\varepsilon<\dfrac{a_{n+1}}{a_n}<\ell+\varepsilon$.
Here is where I have to assume that eventually $a_n>0$ to conclude (by induction) that eventually $$(\ell-\varepsilon)^na_0<a_n<(\ell+\varepsilon)^na_0\\\therefore(\ell-\varepsilon)a_0^{1/n}<a_n^{1/n}<(\ell+\varepsilon)a_0^{1/n}\\\therefore \ell-\varepsilon\leq\lim a_n^{1/n}\leq\ell+\varepsilon\\\therefore\lim a_n^{1/n}=\ell$$ But I haven't yet figured out how to prove this without the extra assumption.

user458811
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    Some assumption of positivity is necessary for this to make sense: otherwise, take $a_n = 1$ when $n$ is odd, $-1$ when even. –  Dec 21 '17 at 04:40
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    Well, if $a_n$ is negative, how do you define, say $\sqrt {a_n}$? Therefore, for the terms $a_n^{\frac 1n}$ to remain real, you would need to ensure that $a_n$ are positive. – Sarvesh Ravichandran Iyer Dec 21 '17 at 04:42
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    That's fine. And the theorem makes sense only for positive $a_n$. – Paramanand Singh Dec 21 '17 at 04:46
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    Note, BTW, that you cannot assert that for every $\epsilon,$ $(l - \epsilon)^n a_0 \le a_n \le \dots$. Indeed, since the $a_{n+1}/a_n \in (l - \epsilon, l + \epsilon)$ only for $n \ge N, $ the best you can say is (modulo off-by-one-errors I haven't checked for) that for each $\epsilon,$ there exists $N$ s.t. for $n > N,$ $(l - \epsilon)^{n - N} a_N \le a_n \le \dots$ Of course, this changes neither the proof strategy nor the conclusion, so you're fine there. – stochasticboy321 Dec 21 '17 at 06:16
  • Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Feb 01 '18 at 22:36

2 Answers2

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Suppose $$a_n >0 \quad\frac{a_{n+1}}{a_n}\to\ell \quad \ell\in [0,+\infty)$$

Thus $\forall\varepsilon>0, \exists n_0\in \mathbb{N}$ s.t. $\forall n\ge n_0$ $$\ell-\frac \varepsilon2\le\frac{a_{n+1}}{a_n}\le\ell+\frac \varepsilon2$$

Therefore

$$\ell-\frac \varepsilon2\le\frac{a_{n_0+1}}{a_{n_0}}\le\ell+\frac \varepsilon2 $$

$$\left(\ell-\frac \varepsilon2 \right)^ka_{n_0}\le a_{n_0+k}\le\left(\ell+\frac \varepsilon2\right)^k a_{n_0}$$

$$\left(\ell-\frac \varepsilon2 \right)^n\left(\ell-\frac \varepsilon2 \right)^{-n_0} a_{n_0}\le a_{n}\le\left(\ell+ \frac \varepsilon2\right)^n \left(\ell+ \frac \varepsilon2\right)^{-n_0} a_{n_0}$$

$$\left(\ell-\frac \varepsilon2 \right)\sqrt[n]{\left(\ell-\frac \varepsilon2 \right)^{-n_0} a_{n_0}}\le \sqrt[n]{a_{n}} \le \left(\ell+ \frac \varepsilon2\right) \sqrt[n]{\left(\ell+ \frac \varepsilon2\right)^{-n_0} a_{n_0}}$$

Thus eventually:

$$\ell-\varepsilon \le \sqrt[n]{a_{n}} \le\ell+ \varepsilon \iff \sqrt[n]{a_{n}}\to \ell$$

Note:

for $\ell=0$ consider:

$$0\le\frac{a_{n_0+1}}{a_{n_0}}\le\ell+\frac \varepsilon2 $$

user
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  • I actually spent some time to overcome and elaborate the "thus definitively" part ;) ... otherwise I don't see why your answer is any different :( – rtybase Dec 23 '17 at 18:14
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    @rtybase I agree with you the proof is almost the same, I've only used a more concise notation and this could make it clearer. – user Dec 23 '17 at 18:23
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To avoid confusions, I will assume

  • $\color{red}{\ell >0}$
  • $a_n >0$ so that $\sqrt[n]{a_n}$ makes sense for all $n$.
  • and $\varepsilon$ mentioned below is $\varepsilon< \ell$.

We are given that $\forall\varepsilon>0, \exists N(\varepsilon)\in \mathbb{N}$ s.t. $\forall n> N(\varepsilon)$ $$\ell-\varepsilon<\frac{a_{n+1}}{a_n}<\ell+\varepsilon \tag{1}$$ Now $$\sqrt[n]{a_n}=\sqrt[n]{\frac{a_{n}}{a_{n-1}}\cdot \frac{a_{n-1}}{a_{n-2}}\cdot ...\cdot \frac{a_{N(\varepsilon)+2}}{a_{N(\varepsilon)+1}} \cdot a_{N(\varepsilon)+1}}$$ or $$\sqrt[n]{a_{N(\varepsilon)+1}}\sqrt[n]{\left(\ell-\varepsilon\right)^{n-N(\varepsilon)-1}} < \sqrt[n]{a_n} <\sqrt[n]{a_{N(\varepsilon)+1}} \sqrt[n]{\left(\ell+\varepsilon\right)^{n-N(\varepsilon)-1}} \iff \\ \left(\ell-\varepsilon\right) \cdot \sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell-\varepsilon)^{N(\varepsilon)+1}}} < \sqrt[n]{a_n} <\left(\ell+\varepsilon\right) \cdot \sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell+\varepsilon)^{N(\varepsilon)+1}}} \tag{2}$$ Both $$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell-\varepsilon)^{N(\varepsilon)+1}}} \rightarrow 1, n\rightarrow\infty$$ $$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell+\varepsilon)^{N(\varepsilon)+1}}} \rightarrow 1, n\rightarrow\infty$$ and from some $n$ onwards (or maximum between it and $N(\varepsilon)$) $$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell-\varepsilon)^{N(\varepsilon)+1}}} > 1-\frac{\varepsilon}{\ell}$$ $$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{(\ell+\varepsilon)^{N(\varepsilon)+1}}} < 1+\frac{\varepsilon}{\ell} \tag{3}$$ or $$\left(\ell-\varepsilon\right) \cdot \left(1-\frac{\varepsilon}{\ell}\right) < \sqrt[n]{a_n} <\left(\ell+\varepsilon\right) \cdot \left(1+\frac{\varepsilon}{\ell}\right)$$ or (because I assumed $\varepsilon< \ell$) $$\ell - 3\varepsilon<\ell-\varepsilon -\varepsilon + \frac{\varepsilon^2}{\ell} < \sqrt[n]{a_n} <\ell+\varepsilon+\varepsilon+\frac{\varepsilon^2}{\ell} < \ell + 3\varepsilon \tag{4}$$ If we "adjust" $(1)$ above to $\frac{\varepsilon}{3}$ then $(4)$ becomes $$\ell - \varepsilon< \sqrt[n]{a_n} < \ell + \varepsilon$$

One obvious question is, what happens when $\color{red}{\ell =0}$?

Then $(1)$ becomes $$0<\frac{a_{n+1}}{a_n}< \varepsilon \tag{1a}$$ Then $(2)$ becomes $$0<\sqrt[n]{a_n} <\varepsilon \cdot \sqrt[n]{\frac{a_{N(\varepsilon)+1}}{\varepsilon^{N(\varepsilon)+1}}} \tag{2a}$$ Then $(3)$ reduces to $$\sqrt[n]{\frac{a_{N(\varepsilon)+1}}{\varepsilon^{N(\varepsilon)+1}}} < 1+\varepsilon \tag{3a}$$ And $(4)$ becomes (assuming $\varepsilon <1$) $$0< \sqrt[n]{a_n} <2\varepsilon \tag{4a}$$ If we "adjust" $(1a)$ above to $\frac{\varepsilon}{2}$ then $(4a)$ becomes $$0< \sqrt[n]{a_n} < \varepsilon$$

rtybase
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