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

  • #1
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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?
 

Answers and Replies

  • #2
StoneTemplePython
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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.
 
Last edited:
  • #3
RPinPA
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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.
 

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