Let f : (0, ∞) → R be a decreasing function. A sequence (xn) is definedby x1 = 1 and xn+1 = xn + f (xn). Show that limn→∞ xn = ∞.

To prove that limn→∞ xn = ∞, we need to show that for every real number M, there exists a natural number N such that for all n > N, xn > M.

Given that f is a decreasing function, we know that for all x > y, f(x) ≤ f(y).

Let's start with the recursive definition of the sequence: xn+1 = xn + f(xn).

Since f is decreasing, we know that f(xn) ≤ f(x1) for all n.

So, xn+1 = xn + f(xn) ≥ xn + f(x1) for all n.

Let's denote f(x1) by a. Then, we have xn+1 ≥ xn + a for all n.

This inequality tells us that each term of the sequence is at least a greater than the previous term.

So, if we choose M to be any real number, we can find a natural number N such that N*a > M (since a is a constant and N can be as large as we want).

Then, for all n > N, we have xn > N*a > M.

Therefore, we have shown that for every real number M, there exists a natural number N such that for all n > N, xn > M.

This is the definition of limn→∞ xn = ∞, so we have proven the statement.