Question

A sequence is defined recursively by the equations \({{\rm{a}}_{\rm{1}}}{\rm{ = 1}}\),\({{\rm{a}}_{{\rm{n + 1}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}\left( {{{\rm{a}}_{\rm{n}}}{\rm{ + 4}}} \right)\)Show that \(\left\{ {{{\rm{a}}_{\rm{n}}}} \right\}\)is increasing and \({{\rm{a}}_{\rm{n}}}{\rm{ < 2}}\), for all n, Deduce that \(\left\{ {{{\rm{a}}_{\rm{n}}}} \right\}\)is convergent and find its limit.

Short answer

Sequence\(\left\{ {{a_n}} \right\}\)is increasing and\({a_n} < 2\)for all\(n\)

The limit of sequence \(\left\{ {{a_n}} \right\},L = 2\)

Step-by-step solution

Definition of convergent

In mathematics, the property of approaching a limit more and more closely as an argument (variable) of the function grows or decreases, or as the number of terms in the series increases (as shown by some infinite series and functions).

Calculating hypothesis n.

Consider the given values and simplify,

\({{\rm{a}}_{{\rm{n + 1}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}\left( {{{\rm{a}}_{\rm{n}}}{\rm{ + 4}}} \right)\)

Let us assume that \(\left\{ {{a_n}} \right\}\)is increasing and \({{\rm{a}}_{\rm{n}}}{\rm{ < 2}}\), then

\({{\rm{a}}_{{\rm{n - 1}}}}{\rm{ < }}{{\rm{a}}_{\rm{n}}}{\rm{ < 2}}\)

We will use mathematical induction method to prove the results.

If \({\rm{n = 2}}\), we have

\(\begin{array}{c}{{\rm{a}}_{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}\left( {{{\rm{a}}_{\rm{1}}}{\rm{ + 4}}} \right)\\{{\rm{a}}_{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{(1 + 4)}}\\{{\rm{a}}_{\rm{2}}}{\rm{ = }}\frac{{\rm{5}}}{{\rm{3}}}\\{{\rm{a}}_{\rm{2}}}{\rm{ < 2}}\\{\rm{ = > 1 < }}{{\rm{a}}_{\rm{2}}}{\rm{ < 2}}\\{\rm{ = > }}{{\rm{a}}_{\rm{1}}}{\rm{ < }}{{\rm{a}}_{\rm{2}}}{\rm{ < 2}}\\\end{array}\)

Therefore, our hypothesis holds for \({\rm{n = 2}}\).

Testing hypothesis for n = (k+1).

Now, for \(n = k\), we have

\({{\rm{a}}_{{\rm{k - 1}}}}{\rm{ < }}{{\rm{a}}_{\rm{k}}}{\rm{ < 2}}\)

Now, we test our hypothesis for \({\rm{n = }}\left( {{\rm{k + 1}}} \right)\). Therefore, we have

\(\begin{array}{c}{{\rm{a}}_{{\rm{k + 1}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}\left( {{{\rm{a}}_{\rm{k}}}{\rm{ + 4}}} \right)\\{{\rm{a}}_{{\rm{k + 1}}}}{\rm{ < }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{(2 + 4)}}\\{{\rm{a}}_{{\rm{k + 1}}}}{\rm{ < }}\frac{{\rm{1}}}{{\rm{3}}}{\rm{(6)}}\\{{\rm{a}}_{{\rm{k + 1}}}}{\rm{ < 2}}\\{\rm{ = > }}{{\rm{a}}_{\rm{k}}}{\rm{ < }}{{\rm{a}}_{{\rm{k + 1}}}}{\rm{ < 2}}\end{array}\)

Our hypothesis holds for \({\rm{n = 1, k}}\) and \({\rm{k + 1}}\). Therefore, our hypothesis is proved to be true by mathematical induction.

Hence, \(\left\{ {{a_n}} \right\}\) is increasing.

And \({{\rm{a}}_{\rm{n}}}{\rm{ < 2}}\) for all \(n\).

Finding whether \({\rm{\{ }}{{\rm{a}}_{\rm{n}}}{\rm{\} }}\) is convergent.

As we have \(1 \le {a_n} < 2\) for all \(n\), therefore any sequence \(\left\{ {{a_n}} \right\}\) will not diverge hence converge.

The limit of the sequence \(\left\{ {{a_n}} \right\}\) is given by

\(\begin{array}{c}L = \mathop {\lim }\limits_{n \to \infty } {a_n}\\L = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}}\\L = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{3}\left( {{a_n} + 4} \right)} \right]\\L = \left[ {\frac{1}{3}(L + 4)} \right]\\3L = L + 4\\2L = 4\\L{\rm{ }} = 2\end{array}\)

As \(L\) exists, \(\left\{ {{a_n}} \right\}\) is convergent.