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Let $a_n$ and $b_n$ two strictly positive sequences such that $$\lim_{n\to\infty}a_n^n=a>0\qquad \lim_{n\to\infty}b_n^n=b>0.$$

I need to prove that $$\lim_{n\to\infty}\left(\frac{a_n+b_n}{2}\right)^n=\sqrt{ab}$$

I have really no idea of what to do and what to look at. Please, could you avoid to give me a complete solution? Could you highlight the things a good mathematician should notice when tackling a problem like this?

Thank you.

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  • $\begingroup$ Isn't that the Babylonian algorithm? $\endgroup$ – K. Rmth Jan 28 '14 at 18:11
  • $\begingroup$ @FalsePromise, placing typographical elements below or above the main line is discouraged in the title, because this often breaks the alignment of the titles on the home page of MSE. $\endgroup$ – Alex M. Dec 30 '16 at 11:43
  • $\begingroup$ @E.Joseph, placing typographical elements below or above the main line is discouraged in the title, because this often breaks the alignment of the titles on the home page of MSE. You should have rejected that edit. $\endgroup$ – Alex M. Dec 30 '16 at 11:43
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Note that $a_n=1+\alpha_n/n$ for some sequence $\{\alpha_n\}$ with $\alpha_n \to \log a$. The same goes for $b_n$ and $\beta_n$. Ergo

$$\lim_{n\to\infty} \Bigl( \frac{a_n+b_n}{2}\Bigr)^n = \lim_{n\to\infty} \Bigl( 1+ \frac{\alpha_n+\beta_n}{2n} \Bigr)^n = \exp\bigl( (\log a + \log b) /2 \bigr) = \sqrt{ab}. $$

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  • $\begingroup$ What can I do to understand why $a_n=1+\alpha_n/n$? I mean can you show me how I can derive it? $\endgroup$ – Charlie Jan 28 '14 at 18:16
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    $\begingroup$ There is always some $\alpha_n$ for which $a_n=1+\alpha_n/n$, correct? Now, use the fact that $(1+t/n)^n \to e^t$. $\endgroup$ – JPi Jan 28 '14 at 18:31
  • $\begingroup$ Thank you, now I can understand. What did suggest you that you should express $a_n=1+\alpha_n/n$? $\endgroup$ – Charlie Jan 28 '14 at 19:09
  • $\begingroup$ The fact that you had a limit for $a_n^n$ and I know the limit for $(1+t/n)^n$. $\endgroup$ – JPi Jan 28 '14 at 19:42
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Bernoulli's Inequality gives $$ \begin{align} 1+n(x-1) &\le(1+(x-1))^n\\[6pt] &=x^n\tag{1} \end{align} $$ For $x\ge1$, apply $(1)$ and $1+x\le e^x$, $$ \begin{align} \left[\frac{1+x}2\frac{1+1/x}2\right]^n &=\left[1+\frac{(x-1)^2}{4x}\right]^n\\ &\le\left[1+\frac{x^{2n}}{4n^2x}\right]^n\\ &\le\exp\left(\frac{x^{2n}}{4nx}\right)\tag{2} \end{align} $$ Factoring out $a_n^n$ and $b_n^n$ and taking the geometric mean $$ \begin{align} \lim_{n\to\infty}\left(\frac{a_n+b_n}2\right)^n &=\lim_{n\to\infty}b_n^n\lim_{n\to\infty}\left[\frac{\frac{a_n}{b_n}+1}2\right]^n\\ &=\lim_{n\to\infty}a_n^n\lim_{n\to\infty}\left[\frac{1+\frac{b_n}{a_n}}2\right]^n\\ &=\sqrt{AB}\lim_{n\to\infty}\left[\frac{\frac{a_n}{b_n}+1}2\frac{1+\frac{b_n}{a_n}}2\right]^{n/2}\\[12pt] &=\sqrt{AB}\tag{3} \end{align} $$ by the Squeeze Theorem and $(2)$.

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I'll try another path: naming $A_n = a_n ^n$ we get $A_n\to a$; similarly for $b_n$ and $B_n$. Now: $$ \lim_{n\to \infty} \left(\frac{a_n+b_n}{2}\right)^n = \lim_{n\to \infty} \left(\frac{A_n^{1/n}+B_n^{1/n}}{2}\right)^n = \left[1^\infty\right] $$ therefore we go for the form $(1+1/n)^{n}\to e$: $$ \lim_{n\to \infty} \left[\left( 1+ \frac{A_n^{1/n}+B_n^{1/n} -2}{2} \right)^{{2}/{(A_n^{1/n}+B_n^{1/n} -2)}}\right]^{p_n} $$ where the expression in the square parenthesis gives $e$ and $$ p_n = n\frac{A_n^{1/n}+B_n^{1/n} -2}{2} = n\frac{e^{(\ln A_n)/n}+e^{(\ln B_n)/n} -2}{2} $$ which yields, with Taylor series expansion: $$ \frac{\ln A_nB_n}{2} + O\left(\frac{1}{n}\right) \to \ln \sqrt{ab}. $$ By collecting the piece we finally have: $$ \lim_{n\to \infty} \left(\frac{a_n+b_n}{2}\right)^n = e^{\ln{\sqrt{ab}}} = \sqrt{ab}. $$

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