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Prove that the sequences $\frac{2a_nb_n}{a_n+b_n}$ and $\sqrt{a_nb_n}$ are both convergent to the same limit for all ai,bi>0. Given the special case of two positive numbers $a_i$ and $b_i$, the harmonic mean would be equal to $\frac{2a_nb_n}{a_n+b_n}$ correct? Could I make a relationship to the geometric mean, $\sqrt{a_nb_n}$?

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  • $\begingroup$ Your question is imprecisely phrased. "Convergence" implies some sort of limiting process. You have specified no such process or structure. You simply write about various means of two numbers without saying what aspect or aspects of these means are to be regarded as a limiting process. $\endgroup$ – heropup Oct 15 '15 at 21:55
  • $\begingroup$ Did you mean to write $\frac{2a_{n-1}b_{n-1}}{a_{n-1}+b_{n-1}}$ and $\sqrt{a_{n-1}b_{n-1}}$? You can typeset this as $\frac{2a_{n-1}b_{n-1}}{a_{n-1}+b_{n-1}}$ and $\sqrt{a_{n-1}b_{n-1}}$. The other possible interpretation of you wrote is $a_n-1$ instead of $a_{n-1}$, similar for $b$. Both these interpretations seem rather unusual. (If it is the first one, why not simply writing $a_n$ instead of $a_{n-1}$?) $\endgroup$ – Martin Sleziak Oct 16 '15 at 4:11
  • $\begingroup$ Although would are assumptions on the sequences $a_n$ and $b_n$. Do you assume that they are convergent? Or do you even assume that these two sequences are convergent to the same limit? (The later seems to me as the most plausible interpretation of your question.) $\endgroup$ – Martin Sleziak Oct 16 '15 at 4:15
  • $\begingroup$ Whatever it is you mean, judging by the forms of these beasts, you probably need to use the inequalities $\frac{a_n+b_n}{2}\geq \sqrt{a_nb_n}\geq \frac{2a_nb_n}{a_n+b_n}$. $\endgroup$ – Hamed Oct 16 '15 at 4:27
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Unless you have some information about $a_n, b_n$ that you have not shared, this is not correct. Let $a_n=1, b_n=2$ for all $n$ and it fails.

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