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Let $\{a_n\}$ be a sequence of positive numbers such that $\lim_{n\to\infty} a_n = L$. Prove that $$\lim_{n\to\infty}(a_1\cdots a_n)^{1/n} = L$$

Proof:

Let $\epsilon > 0$. There exists $N\in\mathbb{N}$ such that if $n\ge N$ then $L-\epsilon < a_n < L + \epsilon$. Now, let $b_n = (a_1\cdots a_n)^{1/n} $. We can split this up based on the tail of $\{a_n\}$: $b_n = (a_1\cdots a_{N})^{1/n} (a_{N+1}\cdots a_{n})^{1/n}$. We can bound $(a_{N+1}\cdots a_{n})^{1/n}$ by other sequence which have limits, since we have that $L-\epsilon < a_n < L + \epsilon$ for all $n\ge N$: $$(L-\epsilon)^{1-N/n} < (a_{N+1}\cdots a_{n})^{1/n} < (L+\epsilon)^{1-N/n}.$$ Let $C=a_1\cdots a_N$. If we multiply this inequality through by $C^{1/n}$ we find that $$C^{1/n}(L-\epsilon)^{1-N/n} < b_n < C^{1/n}(L+\epsilon)^{1-N/n}.$$ If we take the limit of this inequality, we find that $L - \epsilon \le \lim_{n\to\infty} b_n \le L+\epsilon$. So $L \le \lim_{n\to\infty}b_n \le L$. Hence $\lim_{n\to\infty}b_n = L$.

Does this argument work?

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  • $\begingroup$ Yes, this argument is fine. $\endgroup$ – Kavi Rama Murthy Oct 11 '18 at 23:24
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    $\begingroup$ Take logarithm and use Cesaro sequence. $\endgroup$ – hamam_Abdallah Oct 11 '18 at 23:24
  • $\begingroup$ @Murthy Why do some proofs of this (and similarly with the proof that the means) converge use the same argument as mine but instead of taking with lim of the inequality, they separately do limsup and liminf, and show that these are equal? $\endgroup$ – Wesley Oct 11 '18 at 23:47
  • $\begingroup$ Don't bother. Your proof is not only correct it as simple as it can be. $\endgroup$ – Kavi Rama Murthy Oct 12 '18 at 6:39
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    $\begingroup$ Well it is not not known in advance that $\lim b_n$ exists and hence the tactical device of $\limsup, \liminf$ is used. $\endgroup$ – Paramanand Singh Oct 12 '18 at 12:45

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