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Suppose that a sequence ${a_n : n = 1,2,\cdots}$ real numbers is such that $a_n\geq 1$ for all $n ≥ 1$ and $$ a_{n+m}\leq a_na_m\quad \rm{for~ all}~n\geq1, m\geq1. $$ Show that $a_n^{1/n}$ converges as $n\to\infty$

My solution: By taking log, we have

$$ \log{a_{n+m}}\leq \log{a_na_m}=\log{a_n}+\log{a_m}\quad \rm{for~ all}~n\geq1, m\geq1. $$

So we have \begin{align} \log{a_2}&\leq \log{a_1}+\log{a_1}=2\log{a_1} \\ \log{a_3}&\leq \log{a_1}+\log{a_2}\leq3\log{a_1} \\ \cdots \\ \log{a_n}&\leq n\log{a_1} \end{align} So, $$ \log{a_n}^{1/n}=\frac{1}{n}\log{a_n} $$ So I can prove $\log a_n$ is bounded but cannot to prove it's monotonic which can sufficiently lead to $\log a_n$ converges. How to deal with it?

Or if we cannot prove monotonic, how to prove the limit exists?

Gatsby
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1 Answers1

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we have $$a_{n+m}\leq a_na_m\quad $$ taking m=n, we get $$a_{2n}\leq a_n^2\quad $$

taking square roots on both sides, we get

$$ a_{2n}^{1/2}\leq a_n\quad$$

taking $n^{th}$ roots on both sides , we get

$$ a_{2n}^{1/2n}\leq a_n^{1/n}\quad$$

let

$$f(n)=a_n\quad$$

then , we have $$f(2n)\leq f(n)\quad$$

which is a monotic decreasing sequence and hence converges

rahul
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