Need to show sequence divergent?

Question

I have to show that lim a_n = +(infinity) [as n->+(infinity)]
where a_1 >0
and:
a_(n+1) = a_n +(a_n)^(1/2) for all n>=1

It seems to me like it should be really easy, but for some reason I'm struggling.

I proved it is continues, strictly positive, and strictly increasing;

I figure the obvious thing to do is show that it is unbounded, and therefore since it is strictly increasing it diverges to +(infinity), but I'm really struggling to prove that it's unbounded.

Any help either proving it's unbounded above, or just divergent in general would be greatly appreciated!

Answer 1

The easiest way to show this is to show that ∀n, a_n ≥ a₁ + (n-1)*√a₁. This is may be done by induction: first we note that a₁ = a₁ + (1-1)*√a₁, so this holds where n=1. Now, suppose it holds for some n. Then by the definition of a_n, a_(n+1) = a_n + √(a_n) ≥ a₁ + (n-1)*√a₁ + √(a₁ + (n-1)*√a₁) ≥ a₁ + (n-1)*√a₁ + √a₁ = a₁ + ((n+1) - 1)*√a₁, so this holds for a_(n+1). Thus by induction, it holds for all n.

Now just note that a₁ + (n-1)√a₁ is unbounded (this is a direct application of the archimedian property -- just note that for any N, ∃n∈ℕ such that n > N/√a₁), and so the corresponding a_n must be unbounded as well.

Answer 2

I'm not sure how to show this thing is unbounded so I'll try a different approach (proof by contradiction). Let's assume the sequence converges. That imples that as lim n--> infinity (a_n) = lim n--> infinity (a_(n+1)) = x for some nonzero constant x.

We can also use your insight that the sequence is positive and strictly increasing for all n. We can infer from this that the sequence is bounded from below by a_1 (since all subsequent terms are greater than a_1). This will help later.

a_(n+1) = a_n + (a_n)^(1/2)

If we take the limit as n approaches infinity of both sides, this becomes

x = x + (x)^(1/2)
0 = x^(1/2)
x = 0.

Recalling that x = lim n--> infinity (a_n), this suggests that lim n--> infinity (a_n) = 0. However, we know from above that the sequence is bounded from below by a_1 which is some positive number. So lim n--> infinity (a_n) >= a_1. This is a contradiction to the suggestion that lim n--> infinity (a_n) = 0. As such, our assumption that a_n converges is false. Hence, the sequence diverges.

Answer 3

If a_n < 1 then sqrt(a_n) > a_n, so a_(n+1) > 2*a_n, thus the terms double with each step if a_n < 1. So, if a_n < 1 then a_1 was less than 1 and a_n > 2^(n-1)*a_n. This tells you that the term a_(log(base2)[1/a_1]) >1. So for some N, are all n>=N, a_n > 1.

Now once we get past 1, we are all set since we know that if a_n > 1, sqrt(a_n) > 1, so the sequence increase by at least 1 for each term after that. So we know that for any positive integer n, a_(n+N) > n. Thus the sequence is unbounded and diverges!