Recursive Sequences and Convergence
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315 A sequence of terms $u_1, u_2, u_3, \dots$, where $u_1$ is a given positive real number, is defined by $$u_{n+1} = 1 + \frac{1}{u_n}.$$
(i) For the case $u_1 = 1$, write down the values of $u_2$ and $u_3$. [2]
(ii) Show that one of the first three terms of the sequence is close to 1 in each of the following cases:
(a) $u_1$ is very large (e.g. $u_1 = 1\,000\,000$);
(b) $u_1$ is very small (e.g. $u_1 = 10^{-6}$).
[2]
(iii) Find the value of $u_1$ for which $u_2 = u_1$, giving your answer in an exact form. [4]
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Animated Video Solution
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Step by Step Written Solution
In this problem, we are exploring a recursive sequence defined by u sub n plus one equals one plus one over u sub n, where u sub one is a positive real number. Let's tackle it part by part.
Given Sequence
For part one, we are given that u sub one equals one. We need to find u sub two and u sub three.
Part (i): Calculate $u_2$ and $u_3$ for $u_1 = 1$
To find u sub two, we substitute n equals one into the recurrence formula.
This simplifies to one plus one, which is two. So, u sub two equals two.
Next, to find u sub three, we use the value of u sub two.
One plus one half is three halves, or one point five.
Moving to part two, we want to show that one of the first three terms is close to one under different starting conditions. First, consider when u sub one is very large.
Part (ii): Large and Small $u_1$
(a) $u_1$ is very large (e.g., $10^6$)
If u sub one is huge, then one over u sub one is nearly zero. Therefore, u sub two, which is one plus one over u sub one, will be very close to one.
Now consider the case where u sub one is very small.
(b) $u_1$ is very small (e.g., $10^{-6}$)
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