To prove that both sequences a_n and b_n are convergent to the same limit, we will first show that they are both monotonic and bounded. Then, by the Monotone Convergence Theorem, we can conclude that they converge to the same limit.
First, we will prove that a_n is increasing. Since 0<x<y, we can see that a_1=x<y, and a_2=(a_1+b_1)/2 < (y+y)/2 = y. This shows that a_2>a_1. Now, assume that a_n<a_(n+1), then a_(n+1)=(a_n+b_n)/2 > (a_n+a_n)/2 = a_n. Therefore, a_n is increasing.
Next, we will prove that b_n is decreasing. Since 0<x<y, we can see that b_1=y>x, and b_2=sqrt(a_2+b_1) < sqrt(y+y) = y. This shows that b_2<b_1. Now, assume that b_n>b_(n+1), then b_(n+1)=sqrt(a_(n+1)+b_n) < sqrt(a_n+b_n) = b_n. Therefore, b_n is decreasing.
Since a_n is increasing and b_n is decreasing, we can see that a_n<b_n for all n. This means that both sequences are bounded between x and y.
Now, let's find the limit of these sequences. We will use the fact that both a_n and b_n are convergent to the same limit and use the recursive definition of the sequences to solve for this limit.
Let L be the limit of both a_n and b_n. Then, by the recursive definition, we have:
L = (L+L)/2
L = sqrt(L+L)
Solving for L, we get L=0.
Therefore, both sequences a_n and b_n are convergent to the limit L=0. This can also be seen by taking the limit as n approaches infinity of both sequences:
lim(a_n) = lim(b_n) = lim((a_n+b_n)/2) = lim(sqrt(a_n+b_n)) = 0.
In conclusion, we have proved that both sequences a_n and b_n are convergent to the same limit of 0. This is because both sequences are monotonic and bounded, and we have used the
