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