Convergent sequences to the same limit I know that $\sqrt{ab}\leq\frac{a+b}{2}$.
Let $a_{1}$ and $b_{1}$ two positive real numbers such that $a_{1}<b_{1}$. Define $a_{n+1}=\sqrt{a_{n}b_{n}}$ and $b_{n+1}=\frac{a_{n}+b_{n}}{2}$. Show that:
a) $a_{n}\leq b_{n}$ for all $n\geq 1$.  $\checkmark$
b) $\{a_{n}\}$ is increasing and $\{b_{n}\}$ is decreasing.  $\checkmark$
c) $\{a_{n}\}$ and $\{b_{n}\}$ are convergent $\checkmark$ and that tend to the same limit. 
I already can prove a), b), and a half of c), but i don't know how to prove that the sequences tend to the same limit. can you help me? please.
 A: Let $a \stackrel{\rm def}{=} \lim_{n\to\infty} a_n$ and $b \stackrel{\rm def}{=} \lim_{n\to\infty} b_n$, which you have proven exist and satisfy $0<a\leq b$.
Then, by continuity, we have
$$a = \lim_{n\to\infty} a_{n+1} = \lim_{n\to\infty} \sqrt{a_nb_n} = \sqrt{ab}
$$
i.e., $\sqrt{a}=\sqrt{b}$. You can conclude.
I used the limit on $a_{n+1} = \sqrt{a_nb_n}$. You can use the same argument on $b_{n+1} = \frac{a_n+b_n}{2}$ instead, if you prefer, to get $a = \frac{a+b}{2}$, leading to the same conclusion: $a=b$.
A: Notice 
$$b_{n+1} - a_{n+1} = \frac{b_n+a_n}{2} - \sqrt{b_na_n}
= \frac12 (\sqrt{b_n} - \sqrt{a_n})^2 = \frac12\left(\frac{\sqrt{b_n}-\sqrt{a_n}}{\sqrt{b_n}+\sqrt{a_n}}\right)(b_n - a_n)$$
Since your already know $b_n \ge a_n > 0$ for $n \ge 1$, you get
$$
\left|\frac{\sqrt{b_n}-\sqrt{a_n}}{\sqrt{b_n}+\sqrt{a_n}}\right| \le 1 \quad\implies\quad
|b_{n+1} - a_{n+1}| \le \frac12 |b_n - a_n|$$
As a result, for all $n > 1$, you have
$$|b_n - a_n | \le \frac{1}{2^{n-1}}|b_1 - a_1|
\quad\implies\quad \lim_{n\to\infty} ( b_n - a_n ) = 0
\quad\implies\quad \lim_{n\to\infty} b_n = \lim_{n\to\infty} a_n
$$
