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 January 28th, 2011, 10:21 AM #1 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Convergent series -> series of geometric means converges (feel free to edit this and convert my statements to latex. I'm bad at latex, and on my phone...) Claim: if the infinite series a_n converges, and b_n := geometric mean of the first n terms of a, then the infinite series b_n converges. This was "left as an exercise", and my first impulse was to use AM-GM, but the inequality goes in the wrong direction! I was thinking about just taking the square root of each, but they aren't necessarily positive... Besides "use the definitions", I'd appreciate any hints.
 January 28th, 2011, 10:51 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Convergent series -> series of geometric means converges Isn't that "sequence", not "series"? I can't see how (b_n) could be a series.
 January 28th, 2011, 02:54 PM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Convergent series -> series of geometric means converges If you add them all up, then they are a series! b1 = a1 b2 = sqrt(a1*a2) b3 = cuberoot(a1*a2*a3) b4 = ... Sum from n=1 to infinity of b_n converges if sum from n = 1 to infinity of a_n converges. Certainly we could express these statements in terms of the sequences an and bn, but that doesn't suggest a solution ...
 February 3rd, 2011, 10:25 AM #4 Member   Joined: Jan 2011 Posts: 36 Thanks: 0 Re: Convergent series -> series of geometric means converges Perhaps proving the contrapositive, namely "If the sum of b_n diverges then the sum of a_n diverges", might be easier? Just fiddling around with it If $\sum_0^\infty b_i$ diverges then $\lim_{n \rightarrow \infty} b_n \neq 0.$ But $b_n= (\prod_0^n a_i)^{\frac{1}{n}}.$ Then we may write $\lim_{n \rightarrow \infty}\prod_0^n a_i^{\frac{1}{n}} \neq 0.$ Eh, I have to go now. But my intuition is telling me that this might be easier to prove. Will try again later.
February 3rd, 2011, 12:03 PM   #5
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Re: Convergent series -> series of geometric means converges

Quote:
 Originally Posted by GeminiDreams ... If $\sum_0^\infty b_i$ diverges then $\lim_{n \rightarrow \infty} b_n \neq 0.$ ...
I think the converse is true, but not this statement...

 February 3rd, 2011, 01:18 PM #6 Member   Joined: Jan 2011 Posts: 36 Thanks: 0 Re: Convergent series -> series of geometric means converges It must be true since it's the contrapositive of: If $\lim_{n \rightarrow \infty} b_n= 0$ then $\sum_0^\infty b_i$ converges
 February 3rd, 2011, 01:46 PM #7 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Convergent series -> series of geometric means converges But that statement isn't true either! Isn't the harmonic series a counterexample?
February 3rd, 2011, 02:42 PM   #8
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Re: Convergent series -> series of geometric means converges

Quote:
 Originally Posted by GeminiDreams It must be true since it's the contrapositive of: If $\lim_{n \rightarrow \infty} b_n= 0$ then $\sum_0^\infty b_i$ converges
That statement, like its contrapositive, is false. The harmonic series and the sum of the reciprocals of the primes are the most famous counterexamples.

 February 6th, 2011, 10:07 AM #9 Member   Joined: Jan 2011 Posts: 36 Thanks: 0 Re: Convergent series -> series of geometric means converges My mistake. Just to be clear. Is this statement equivalent to the question? If $\lim_{n \rightarrow \infty} a_n= C_1$ for $C_1 \in \R$ then $\sum_{n=0}^\infty b_n = C_2$ for $C_2 \in \R$ where $b_n= \left(\prod_{n=0}^\infty a_n\right)^{\frac{1}{n}}$
February 6th, 2011, 10:18 AM   #10
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Re: Convergent series -> series of geometric means converges

Quote:
 Originally Posted by GeminiDreams My mistake. Just to be clear. Is this statement equivalent to the question? If $\lim_{n \rightarrow \infty} a_n= C_1$ for $C_1 \in \R$ then $\sum_{n=0}^\infty b_n = C_2$ for $C_2 \in \R$ where $b_n= \left(\prod_{i=0}^n a_I\right)^{\frac{1}{n}}$
That, except with a lowercase "i" as the indexing subscript of a

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