Problem 13 *

I think the poster of this question missed a subtlety that we need to prove that this expression represents a finite value first. With this caveat - it becomes an early exercise in real analysis.

Define the sequence \langle a_n\rangle by:

a_{n + 1} = \sqrt {42 + a_n}

with a_0 = \sqrt {42}.

We want to show that \langle a_n\rangle converges. (the question being understood as the limit of this sequence) We do this by proving that \langle a_n\rangle is monotone and bounded.

To prove it is bounded - note that a_0 = \sqrt {42} < \sqrt {49} = 7. Suppose that a_n < 7 for any n. Then a_{n + 1} = \sqrt {42 + a_n} < \sqrt {42 + 7} = \sqrt {49} = 7. So a_n < 7 for all n by induction. Clearly we also have 0 < a_n < 7 by the definition of the square root.

To prove that it is monotone, note that:

a_{n + 1} - a_n = \sqrt {42 + a_n} - a_n

Then (since a_n > 0) we have a_{n + 1} > a_n iff 42 + a_n > a_n^2. That is, a_n^2 - a_n - 42 = (a_n - 7)(a_n + 6) < 0.

Since 0 < a_n < 7, we are assured this is the case. So \langle a_n\rangle is monotone increasing. Since it’s also bounded, we have a_n \to l \in \mathbb R (with 0 < l \le 7 since 0 < a_n < 7). We therefore also have a_{n + 1} \to l and a^2_{n + 1} \to l^2 from the product rule. So from the relation:

a_{n + 1}^2 = 42 + a_n

taking n \to \infty we have:

l^2 = 42 + l

so:

(l - 7)(l + 6) = 0

Since l > 0 we have l = 7. That is:

\sqrt{42 + \sqrt {42 + \sqrt {42 + \ldots}}} = 7

If we replace 42 with 6, the process is identical showing that the sequence converges, and we would instead have l^2 - l - 6 = (l + 2)(l - 3) = 0 and from the bounds we’d obtain l = 3.For 2 we’d have l^2 - l - 2 = (l - 2)(l + 1) = 0 and hence l = 2. For 8 we’d have l^2 - l - 8 = 0 and so l = \dfrac {1 + \sqrt {33}} 2.

I’m not sure what they mean by “work”.

Do they mean for which the sequence converges?

At the level this is pitched at, this would mean that l^2 - l - a = 0 has no positive solutions, which requires both:

\displaystyle \frac 1 2 \left(1 - \sqrt {4a + 1}\right) \ge 0

and:

\displaystyle \frac 1 2 \left(1 + \sqrt {4a + 1}\right) \ge 0

It suffices that:

1 \ge \sqrt {4a + 1}

ie that a \ge 0. (which was implied anyway)

The sequence converges to an integer?

Having a = n(n - 1) for some positive integer n suffices. (since the roots of l^2 - l - a = 0 necessarily differ by 1)

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These are lifted from that TSR doc so looks like there will be some correcting required of questions. Additionally, not sure if there is a question ‘1’