We have that $0<b\le a$ are two positive real numbers. Their arithmetic mean is defined as $a_1=(a+b)/2$ and their geometric mean is $b_1=\sqrt{ab}$.
Firstly we have to show an that $b_1\le a_1$ iff $a=b$, so we must show both directions algebraically.
Then we define two sequences by setting $a_{n+1}=(a_n+b_n)/2$ and $b_{n+1} = \sqrt{a_nb_n}$. Using mathematical induction to show : $a\ge a_1\ge\dots \ge a_n\ge b_n\ge\dots\ge b_1\ge b$.
Prove that both series converge and show their limit.
I'm having trouble with the first step in proving the iff statement in the forward and backward direction, as well as showing the proof by induction. Also how would one determine the convergence and limit of this series?
First, if $a=b$ then $$ b_1=a=a_1 $$ Now, let $a=1$ and $b=2$. Then $$ a_1=1.5 \ \text{ and }b_1=\sqrt{2} \approx 1.414 $$ Hence $a \ne b$ and $b_1 \leq a_1$. There's a problem with what you wrote isn't it ?
It might be $a_1 \leq b_1$ iff $a=b$. Indeed, suppose $$ a_1 \leq b_1 $$ Then using the fact that $x \mapsto x^2$ is increasing $$ a+b \leq 2\sqrt{ab} \Rightarrow a^2+2ab+b^2 \leq 4ab \Rightarrow \left(a-b\right)^2 \leq 0 $$ With the last inequality, we clearly see that the only possibility is $a=b$. ( which justifies directly the iff )
So in other term, we ALWAYS have $b_1 \leq a_1$.
The sequences you mention are well-known and closely linked to elliptical integrals ( discovery of Abel and Bernoulli ) because it is also linked with the length of the lemniscate. You can show that for $a=1$ and $b=\sqrt{2}$ both $(a_n)$ and $(b_n)$ converge to the same limit which is the length of a lemniscate of Bernoulli of parameter $1$ ( if i remember well ).
You can prove it by induction, but you can wisely use the property of "adjacent" sequences. https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_des_suites_adjacentes ( sorry it's french, does not know the name in english ).
You can show that $(a_n)$ increases, $(b_n)$ decreases and $a_n-b_n \underset{n \rightarrow +\infty}{\rightarrow}0$. In fact, they both converges to the same limit which is often wrote as $M\left(a,b\right)$ that stands for arithmetico-geometric mean.
You cannot find explicitly the value of the common limit $M\left(a,b\right)$. However for fun, you have beautiful properties $$ -M\left(a,b\right)=M\left(b,a\right) $$ $$ -M\left(\alpha a,\alpha b\right)=\alpha M\left(a,b\right) $$ $$ \text{min}\left(a,b\right) \leq \sqrt{ab} \leq M\left(a,b\right) \leq \frac{a+b}{2} \leq \text{max}\left(a,b\right) $$ And last but not least $$ \frac{2}{\pi}M\left(a,b\right)=\int_{0}^{\pi/2}\frac{\text{d}t}{\sqrt{x^2\cos^2(t)+y^2\cos^2(t)}} $$