- #1

Mr Davis 97

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## Homework Statement

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

## Homework Equations

## The Attempt at a Solution

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}## 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 limsup of this inequality, we find that ##L \le \limsup_{n\to\infty} b_n \le L##. So ##\limsup_{n\to\infty}b_n = L##. Similarly, ##\liminf_{n\to\infty}b_n = L##. So ##\lim_{n\to\infty}b_n = L##.

Does this argument work? Could this be proved without the use of limit superior or inferior, i.e. directly with just the definition of convergence known?