a Problem about Special Sequence I am doing homework. And this question may related to #Proving statement about sequences. However, I want to consider the limit of it. Let we assume $a_1=a,b_1=b,b>a>0$. Then we consider the next three iterations: $$\tag1a_{n+1}=\sqrt{a_nb_n},b_{n+1}=\frac{a_n+b_n}{2}$$ $$\tag2a_{n+1}=\frac{2}{\frac{1}{a_n}+\frac{1}{b_n}},b_{n+1}=\frac{a_n+b_n}{2}$$ $$\tag3a_{n+1}=\frac{2}{\frac{1}{a_n}+\frac{1}{b_n}},b_{n+1}=\sqrt{a_nb_n}$$ Now we denote three limits are $AG(a,b),AH(a,b),GH(a,b)$. The question is to represent the limit by use the symbol $$I=\int_{0}^{\frac{\pi}{2}}\frac{dx}{\sqrt{a^2\sin^2x+b^2\cos^2x}}$$ > **Attemptation** It is easy to compute the $AH(a,b)=\sqrt{ab}$. But I am confused with the other two: why is it related to $I$. I can not feel it.

Yes, $(2)$ is easy once you notice that $a_{n+1}b_{n+1}=a_nb_n$, so the product (and hence also the geometric mean) is invariant. As it is well-know that it is between the harmonic and arithmetic mena, the limit of our nested intervals must equal $\sqrt{ab}$.

A similar approach works for $(1)$: Show that $I(a,b)=I(\sqrt{ab},\frac{a+b}2)$. Indeed, $$ I\left(\sqrt{ab},\frac{a+b}2\right)=\int_0^{\frac\pi2}\frac{\mathrm dx}{\sqrt{ab\sin^2x+\frac14(a^2+2ab+b^2)\cos^2x}}$$ can be nicely attacked by trigonometirc identities (and - like the geometric mean above - we have that a simple expression in $I(a,b)$ is between $a$ and $b$).