Define (xn)n≥1 by x1 := 0 and xn+1 := (xn + 3)/4 for n ≥ 1. Prove that the sequence (xn)n≥1 isconvergent and calculate the limit limn→∞ xn

To prove that the sequence (xn)n≥1 is convergent, we can use the Monotone Convergence Theorem. This theorem states that every bounded and monotone sequence is convergent.

Step 1: Prove that the sequence is increasing (monotone)

We can prove this by induction. For n=1, x1=0 and x2=(x1+3)/4=3/4. So, x2>x1.

Assume that xn+1>xn for some n. We need to prove that xn+2>xn+1.

xn+2 = (xn+1 + 3)/4 = (xn + 3 + 3)/4 = (xn + 6)/4 > (xn + 3)/4 = xn+1.

So, the sequence is increasing.

Step 2: Prove that the sequence is bounded

We can also prove this by induction. For n=1, x1=0 which is less than or equal to 1.

Assume that xn<=1 for some n. We need to prove that xn+1<=1.

xn+1 = (xn + 3)/4 <= (1 + 3)/4 = 1.

So, the sequence is bounded.

By the Monotone Convergence Theorem, the sequence (xn)n≥1 is convergent.

To calculate the limit, we set the limit of the sequence equal to a variable L.

limn→∞ xn = L

Since the sequence is convergent, we can set up the following equation:

L = (L + 3)/4

Solving for L gives L = 1.

So, the limit of the sequence as n approaches infinity is 1.