If lim a_n = L, then the geometric mean converges to L

In summary, the conversation discusses the proof for the limit of a sequence of positive numbers. The proof utilizes the concept of Cesaro mean and the equation ##\text{GM} \leq \text{AM}## to show that the limit of the sequence is equal to the limit of the arithmetic mean. The proof also considers the special case where the limit is zero and explains how to handle it. However, the proof requires making both the limiting process and the choice of epsilon explicit to fully understand it.
  • #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?
 
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  • #2
What are your relevant equations?

Do you have access to logarithms or exponential maps or ##\text{GM} \leq \text{AM}##?

The upper bound should be obvious by ##\text{GM} \leq \text{AM}## and the fact that if a limit of a sequence exists, then you get it from the Cesaro mean which is what the limiting value of ##\text{AM}## gives you. The lower bound is a bit trickier and requires a way of bounding the defect in ##\text{GM} \leq \text{AM}##.

Note that a product of non-negative terms is always non-negative, so the above immediately address the special case of ##L=0## (i.e. the GM is squeezed between 0 and 0). For all other cases you could divide by ##L## / WLOG assume that the limit is one, which should be nice to work with since it is the identity element for products. Again, relevant equations seem to be needed so we know what you know and don't.
 
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  • #3
Mr Davis 97 said:
Let ##\epsilon > 0.## There exists ##N\in\mathbb{N}## such that if ##n\ge N## then ##L-\epsilon < a_n < L + \epsilon.##

So from this point forward in the proof, we have picked a particular value of ##\epsilon## which will now remain fixed, and that gives rise to a particular ##N##.

But then I can't see what's happening with that limit. If I fix ##\epsilon## at 0.1 say, what is your argument that ##\limsup_{n \rightarrow \infty} {C^{1/n} (L - 0.1)(L - 0.1)^{-N/n}}## is L and not L - 0.1?

I think you're taking an additional limiting process of letting ##\epsilon## become arbitrarily small, repeating the limit as ##n \rightarrow \infty## for each choice of ##\epsilon## and in that case I believe a simple sandwich argument will work. But I think you have to make BOTH your limiting processes explicit.
 

Related to If lim a_n = L, then the geometric mean converges to L

What does the statement "If lim a_n = L, then the geometric mean converges to L" mean?

The statement means that if the limit of the sequence a_n approaches a specific value L, then the geometric mean of the terms in the sequence will also approach L.

What is a sequence and how is it related to the geometric mean?

A sequence is a list of numbers that follow a specific pattern. The geometric mean is a type of average that is calculated by taking the nth root of the product of n numbers in a sequence.

How does the geometric mean compare to other types of averages?

The geometric mean considers the relative size of the terms in a sequence, while other averages such as the arithmetic mean only consider their sum. This makes the geometric mean useful for analyzing data with exponential growth or decay.

What types of sequences will the geometric mean converge to a specific value?

If the limit of the sequence a_n approaches a specific value L, then the geometric mean will converge to L as long as the terms in the sequence are all positive and the sequence is not decreasing too quickly.

Can the geometric mean diverge or approach a value other than the limit of the sequence?

Yes, the geometric mean can diverge or approach a different value if the terms in the sequence are not all positive or if the sequence is decreasing too quickly. In these cases, the geometric mean may not accurately represent the behavior of the sequence.

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