Question

(a) A sequence is defines recursively by the equation \({a_n} = \frac{1}{2}\left( {{a_{n - 1}} + {a_{n - 2}}} \right)\)for \(n \ge 3\), where \({a_1}\)and \({a_2}\) can be any real numbers/ Experiment with various values of \({a_1}\)and \({a_2}\) and use your calculator to guess the limit of the sequence.

(b) Find \(\mathop {\lim }\limits_{n \to \infty } {a_n}\) in terms of \({a_1}\)and \({a_2}\)by expressing \({a_{n + 1}} - {a_n}\) in terms of \({a_2} - {a_1}\) and summing a series.

Short answer

(a) The series may be tending towards\(\frac{2}{3}\).

(b) The value of summing a series is given by\(\sum\limits_{n = 1}^\infty {{d_n}} = \frac{2}{3}\left( {{a_2} - {a_1}} \right)\)

Step-by-step solution

Same values for \({a_1}\)and \({a_2}\)

Consider a sequence that satisfies the equation \({a_n} = \frac{1}{2}\left( {{a_{n - 1}} + {a_{n - 2}}} \right)\), and numbers \({a_1}\)and \({a_2}\).

Let’s experiment with some initial values by taking the values \({a_1} = 1\)and \({a_2} = 1\).

Rewriting the equation with these values

\( {a_1} = 1\)

\( {a_2} = 1\)

\( {a_3} = \frac{1}{2}\left( {1 + 1} \right)\)

\( {a_3} = 1\)

\( {a_4} = \frac{1}{2}\left( {1 + 1} \right)\)

\( {a_4} = 1\)

Since the next term is always the average of the previous two values, if we start with two equal values the sequence will always be constant.

Different values for \({a_1}\)and \({a_2}\)

Let’s experiment with some initial values by taking the values \({a_1} = 0\)and \({a_2} = 1\).

\({a_1} = 0\)

\({a_2} = 1\)

\( {a_3} = \frac{1}{2}\left( {0 + 1} \right)\)

\( {a_3} = \frac{1}{2}\)

\( {a_4} = \frac{1}{2}\left( {1 + \frac{1}{2}} \right)\)

\( {a_4} = \frac{3}{4}\)

\( {a_5} = \frac{1}{2}\left( {\frac{1}{2} + \frac{3}{4}} \right)\)

\( {a_5} = \frac{5}{8}\)

\( {a_6} = \frac{1}{2}\left( {\frac{5}{8} + \frac{3}{4}} \right)\)

\( {a_6} = \frac{{11}}{{16}}\)

\( {a_7} = \frac{1}{2}\left( {\frac{5}{8} + \frac{{11}}{{16}}} \right)\)

\( {a_7} = \frac{{21}}{{32}}\)

\( {a_8} = \frac{1}{2}\left( {\frac{{21}}{{32}} + \frac{{11}}{{16}}} \right)\)

\( {a_8} = \frac{{43}}{{64}}\)

\( {a_9} = \frac{1}{2}\left( {\frac{{43}}{{64}} + \frac{{21}}{{32}}} \right)\)

\( {a_9} = \frac{{85}}{{128}}\)

\( {a_{10}} = \frac{1}{2}\left( {\frac{{85}}{{128}} + \frac{{43}}{{64}}} \right)\)

\( {a_{10}} = \frac{{171}}{{256}}\)

\( {a_{11}} = \frac{1}{2}\left( {\frac{{85}}{{128}} + \frac{{171}}{{256}}} \right)\)

\({a_{11}} = \frac{{341}}{{512}}\)

Hence,\({a_{11}}\) is equal to the value\(0.66601...,\)approximately.

And, previous values are near this so it will appear that the series may be tending towards\(\frac{2}{3}\).

Calculate the value of \(\mathop {\lim }\limits_{n \to \infty } {a_n}\)

Let \({d_n} = {a_{n + 1}} - {a_n}\) and \({s_n}\) be the partial sum of \({d_n}\).

\({s_n} = {a_2} - {a_1} + {a_3} - {a_2} + {a_4} - {a_3} + ...... + {a_{n + 1}} - {a_n}\)

\({s_n} = {a_{n + 1}} - {a_1}\)

And, the summation is given by

\(\sum\limits_{n = 1}^\infty {{d_n}} = \mathop {\lim }\limits_{n \to \infty } {s_n}\)

\( = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}} - {a_1}\)

This tells us that\(\mathop {\lim }\limits_{n \to \infty } {a_n} = {a_1} + \sum\limits_{n = 1}^\infty {{d_n}} \)

Here, we need to find a formula \(\sum\limits_{n = 1}^\infty {{d_n}} \)

\({a_{n + 1}} - {a_n} = \frac{1}{2}\left( {{a_n} + {a_{n - 1}}} \right) - \frac{1}{2}\left( {{a_{n - 1}} + {a_{n - 2}}} \right)\)

\( = \frac{1}{2}\left( {{a_n} - {a_{n - 2}}} \right)\)

\( = \frac{1}{2}\left( {\frac{1}{2}\left( {{a_{n - 1}} + {a_{n - 2}}} \right) - {a_{n - 2}}} \right)\)

\( = \frac{1}{2}\left( {\frac{1}{2}{a_{n - 1}} - \frac{1}{2}{a_{n - 2}}} \right)\)

\( = \frac{1}{4}\left( {{a_{n - 1}} - {a_{n - 2}}} \right)\)

\({d_n} = \frac{1}{4}{d_{n - 2}}\)

We can get all values in terms of \({d_1}\) and \({d_2}\).

\(\sum\limits_{n = 1}^\infty {{d_n}} = {d_1} + {d_2} + {d_3} + {d_4} + {d_5} + {d_6} + {d_7} + ....\)

\(= {d_1} + {d_2} + \frac{1}{4}{d_1} + \frac{1}{4}{d_2} + \frac{1}{4}\left( {\frac{1}{4}{d_1}} \right) + \frac{1}{4}\left( {\frac{1}{4}{d_2}} \right) + ....\)

\( = {\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_1} + {\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_2}\)

Here, we have tobreak the series into two convergent seriesbecause

\({\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_1}\) and \({\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_2}\) are convergent geometric series.

So, their sum \({\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_1} + {\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_2} = {d_1} + {d_2} + \frac{1}{4}{d_1} + \frac{1}{4}{d_2} + \frac{1}{4}\left( {\frac{1}{4}{d_1}} \right) + \frac{1}{4}\left( {\frac{1}{4}{d_2}} \right) + ....\) is

convergent and \({\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_1} + {\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_2} = {\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_1} + {\left( {\frac{1}{4}} \right)^{n - 1}}{d_2}\)

Thus,

\({\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_1} + {\sum\limits_{n = 1}^\infty {\left( {\frac{1}{4}} \right)} ^{n - 1}}{d_2} = \frac{{{d_1}}}{{1 - \left( {\frac{1}{4}} \right)}} + \frac{{{d_2}}}{{1 - \left( {\frac{1}{4}} \right)}}\)

\( \Rightarrow \frac{4}{3}\left( {{d_1} + {d_2}} \right)\)

Substitute the values of and

We have the values of d₁ = a_{2 -}-a₁and d₂=a₃-a₂

\( = \frac{4}{3}\left( {{a_2} - {a_1} + {a_3} - {a_2}} \right)\)

\( = \frac{4}{3}\left( {{a_3} - {a_1}} \right)\)

\( = \frac{4}{3}\left( {\frac{1}{2}({a_1} + {a_2}) - {a_1}} \right)\)

\( = \frac{2}{3}\left( {{a_2} - {a_1}} \right)\)