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Let $\{a_n\}$ be a sequence of positive numbers such that $\lim_{n\to \infty} a_n = L$. Prove that $$\lim_{n\to \infty} (a_1a_2\ldots a_n)^{\frac 1n} = L$$

(Also given HINT: Let $a>0$. Then $\lim_{n\to\infty} a^{\frac 1n}=1$) What I have so far: $$\text{Let}\:\epsilon>0.\:\text{Then}\;\exists\;N\in\mathrm{I}\!\mathrm{N}\;\text{such that}\;n\ge N.$$$$\text{Then}\;L-\epsilon\lt a_n \lt L + \epsilon$$ $$\text{Let}\; b_n =(a_1a_2\ldots a_n)^{\frac 1n} \text{for}\; n\ge N$$ $$\text{Now}\;b_n = (a_1a_2\ldots a_N)^{\frac 1n}(a_{N+1}\ldots a_n)^{\frac 1n}$$ I am struggling with how to use the given hint and maybe limit superior/inferior facts to prove the limit is L. Any tips would be greatly appreciated. Also this is my first time using MathJax so please let me know if I have made any mistakes. Thank you.

EDIT: I'm looking specifically on how to prove it using $\limsup$ or $\liminf$ properties!

Masacroso
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Jokus
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    Take logs.$\phantom{oooooo}$ – anomaly Jan 16 '18 at 19:03
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    The other question is going for deletion, what happens if you close a duplicate and the other question is deleted? – clark Jan 16 '18 at 19:05
  • Use that $(a_1a_2\ldots a_N)^{\frac{1}{n}}=(a_1)^{\frac{1}{n}}(a_2)^{\frac{1}{n}}\cdots(a_N)^{\frac{1}{n}}$ and that each term goes to 1. – clark Jan 16 '18 at 19:08
  • If each term goes to 1, how does the limit approach L? – Jokus Jan 16 '18 at 19:12
  • @clark: Wait, are you claiming that $\lim_{n\to\infty} \prod_{i=1}^n c_i = \prod_{i=1}^\infty (\lim_{n\to\infty} c_i)$? – anomaly Jan 16 '18 at 19:12
  • @anomaly, jokus I am using the notation that the OP is using, there $N$ is a fixed number. So the first $N$ terms go to 1 and the others will go between $L-\epsilon$ and $L+\epsilon$ – clark Jan 16 '18 at 19:14
  • set $x_n:=\prod_{k=1}^na_k$. Now try to prove that if $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}$ exists then it is equal to $\lim_{n\to\infty}\sqrt[n]{x_n}$. Take a look here also. – Masacroso Jan 16 '18 at 19:23
  • @Masacroso IMO this is a harder problem to do , basically a version of the Cesaro Stoltz lemma. So, if the OP hasn't managed to solve this problem I think it would unlikely to solve the one you are suggesting. – clark Jan 16 '18 at 19:24
  • Juko, each term goes to $1$ but you have infinite terms, which gives the indeterminate form $1^{\infty}$, which in turn can be anything – clark Jan 16 '18 at 19:31
  • See this also https://math.stackexchange.com/questions/2441672/show-that-lim-n-to-infty-sqrt-sum-limitsk-0n-lambda-k-prod-limits – Guy Fsone Jan 16 '18 at 20:05

4 Answers4

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With the ideas of limsup and liminf in mind, we can conjure up an alternate approach than given in other answers.

Let $\{a_n\}$ be a positive sequence such that $a_n\rightarrow L$. Assume, to start, that $L>0$. Choose $\varepsilon$ such that $L>\varepsilon>0$. Then choose $N$ large enough so that for all $n>N$, we have that $$L-\varepsilon\leq a_n\leq L+\varepsilon\,.$$

Finally, set $M=a_1\ldots a_N$ to save on space. Then we have that $$\sqrt[n]{M}\sqrt[n]{(L-\varepsilon)^{n-N}}\leq\sqrt[n]{a_1\ldots a_n}=\sqrt[n]{Ma_{N+1}\ldots a_n}\leq \sqrt[n]{M}\sqrt[n]{(L+\varepsilon)^{n-N}}\,.$$

The inequalities on the left and right respectively give $$L-\varepsilon\leq \liminf\sqrt[n]{a_1\ldots a_n}\quad\text{ and }\quad \limsup\sqrt[n]{a_1\ldots a_n}\leq L+\varepsilon\,.$$

Since $\varepsilon$ is arbitrary, we get our desired limit.

If $L=0$, we needn't worry about the inequality on the left. We can settle for the cruder estimate $$0\leq \sqrt[n]{a_1\ldots a_n}$$ while still using liminf.

  • How do we go from $L-\varepsilon$ to $\sqrt[n]{M}\sqrt[n]{(L-\varepsilon)^{n-N}}$ ? I don't really understand where n-N comes from – Jokus Jan 16 '18 at 23:26
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    The second line of inequalities is self-contained. I used that $(L-\varepsilon)\leq a_{k}$ for $k=N+1, N+2, \ldots, n$ (for a total of $n-N$ times). The $M$ is just along for the ride. –  Jan 16 '18 at 23:29
  • Hi sorry for another late question. For the L=0 case, why don't we worry about the left inequality? Don't we have to show $\liminf \ge 0 ;$ using left inequality, and then use the right inequality to show $\limsup \le 0$? – Jokus Jan 18 '18 at 19:03
  • @Jokus We do worry about the left side, but we use the weaker inequality. $L-\varepsilon$ might be negative, in which case the roots don't make sense. And we use the right inequality to show that $\limsup \leq \varepsilon$ for every positive $\varepsilon$. So, essentially $\limsup\leq 0$, but not directly. –  Jan 18 '18 at 20:17
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Note that

$$(a_1a_2\ldots a_n)^{\frac 1n}=e^{\frac{\sum\log a_i}{n}}$$

and by Stolz-Cesaro

$$\lim_{n\to \infty}\frac{\sum_i^n\log a_i}{n}=\lim_{n\to \infty} \frac{\sum_i^{n+1}\log a_i-\sum_i^{n}\log a_i}{n+1-n}=\lim_{n\to \infty} \log a_{n+1}=\log L$$

thus

$$\lim_{n\to \infty} (a_1a_2\ldots a_n)^{\frac 1n} = e^{\log L}=L$$

user
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The trick is to convert the geometric mean to an arithmetic mean. See here . Then find the limit of the arithmetic mean. See here.

Andrew
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  • it is a particular case of this https://math.stackexchange.com/questions/2441672/show-that-lim-n-to-infty-sqrt-sum-limitsk-0n-lambda-k-prod-limits – Guy Fsone Jan 16 '18 at 20:05
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First, for all $\epsilon>0$ there exists $n_0\in\mathbb{N}$ such that for all $n\geq n_0$ we have $$ L-\epsilon< a_n< L+\epsilon\quad(1) $$ We know that $$ \frac{n}{\frac{1}{a_1}+\cdots+\frac{1}{a_n}}\leq(a_1a_2\cdots a_n)^{\frac{1}{n}}\leq\frac{a_1+\cdots+a_{n}}{n}\quad(2) $$ Hence, we have $$ \frac{n}{\frac{1}{a_1}+\cdots+\frac{1}{a_{n_0}}+\frac{(n-n_0)}{(L-\epsilon)}}\leq(a_1a_2\cdots a_n)^{\frac{1}{n}}\leq\frac{a_1+\cdots+a_{n_0}}{n}+\frac{(n-n_0)}{n}(L+\epsilon)\quad(3) $$ Note that $$ \lim_{n\to\infty}\left[\frac{a_1+\cdots+a_{n_0}}{n}+\frac{(n-n_0)}{n}(L+\epsilon)\right]=L+\epsilon $$ and $$ \lim_{n\to\infty}\frac{\frac{1}{a_1}+\cdots+\frac{1}{a_{n_0}}+\frac{(n-n_0)}{(L-\epsilon)}}{n}=\frac{1}{L-\epsilon} $$ Then for all $\epsilon>0$ $$ L-\epsilon\leq\lim_{n\to\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}}\leq L+\epsilon $$

VJunior
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