# Proof of $\lim \limits_{n\to\infty}\left(\frac{a_n+b_n}{2}\right)^n=\sqrt{ab}$

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.

• Isn't that the Babylonian algorithm? – K. Rmth Jan 28 '14 at 18:11
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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
$$\lim_{n\to\infty} \Bigl( \frac{a_n+b_n}{2}\Bigr)^n = \lim_{n\to\infty} \Bigl( 1+ \frac{\alpha_n+\beta_n}{2n} \Bigr)^n = \exp\bigl( (\log a + \log b) /2 \bigr) = \sqrt{ab}.$$
• What can I do to understand why $a_n=1+\alpha_n/n$? I mean can you show me how I can derive it? – Charlie Jan 28 '14 at 18:16
• There is always some $\alpha_n$ for which $a_n=1+\alpha_n/n$, correct? Now, use the fact that $(1+t/n)^n \to e^t$. – JPi Jan 28 '14 at 18:31
• Thank you, now I can understand. What did suggest you that you should express $a_n=1+\alpha_n/n$? – Charlie Jan 28 '14 at 19:09
• The fact that you had a limit for $a_n^n$ and I know the limit for $(1+t/n)^n$. – JPi Jan 28 '14 at 19:42
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 $(2)$.
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: $$\lim_{n\to \infty} \left(\frac{a_n+b_n}{2}\right)^n = e^{\ln{\sqrt{ab}}} = \sqrt{ab}.$$