Recursive definitions of terms of sequences

Another way to specify the terms of a sequence is using a recursive definition. In a recursive definition, the first term or so is given. Then a formula is given describing how to generate other terms of the sequence from terms already known.

For example, \(c_1 = 2,\quad c_{n+1} = 3c_n – 1\) recursively defines the sequence \(2, 5, 14, 41, 122, \cdots\)

Exercise 8.1.3

Write the first five terms of the sequence defined recursively by \(c_1 = 1,\quad c_2 = 1,\quad c_{n+2} = c_n + c_{n+1}\).

This is a famous sequence called the Fibonacci Sequence.

See Solution

The limit of a sequence

An important issue in Calculus is whether or not a given infinite sequence converges to a limit.

Consider the two infinite sequences

\(1, 2, 3, 4, 5, \cdots = \{ n \}\)

and

\(0.9,~0.99,~0.999,~0.9999,~0.99999, \cdots = \{ 1 – ( 0.1 )^n \}\)

The terms of the first sequence get larger and larger, without any upper bound, whereas the terms of the second sequence have an upper bound. Any number larger than or equal to \(1\) is an upper bound on the terms of the second sequence. The least upper bound of the second sequence is \(1\). We say that the second sequence converges to a limit of \(1\), whereas the first sequence diverges.

A tail of an infinite sequence \(c_1, c_2, c_3, \cdots\) is a sequence \(c_p, c_{p+1}, c_{p+2}, \cdots\) for \(p\ge1\).

For example, \(c_5, c_6, c_7, \cdots\) is a tail of the sequence \(c_1, c_2, c_3, \cdots\)

The statement that the infinite sequence \(c_1, c_2, c_3, \cdots\) converges to the limit \(L\) means that any open interval \((a,b)\) containing \(L\) contains a tail of the sequence.

A arrow is sometimes used to denote convergence. Thus \(c_1, c_2, c_3, \cdots\to L\) means that the sequence converges to the limit \(L\).

To see that the sequence \(\{ c_n \} = \{ 1 – ( 0.1 )^n \}\) converges to a limit of \(1\), suppose \(( a, b )\) is an open interval containing \(1\).

Then \(a < 1\). And no matter how close \(a\) is to \(1\), there is a positive integer \(p\) such that \(( 0.1 )^p\) is smaller than \(1 – a\).

Then \(a\) is smaller than \(1 – ( 0.1 )^p\), which is less than \(1 – ( 0.1 )^{p+1}\) , etc.

But these are terms of the tail \(c_p, c_{p+1}, c_{p+2}, \cdots\) of \(\{ 1 – ( 0.1 )^n \}\).

So all the terms of the tail \(c_p, c_{p+1}, c_{p+2}, \cdots\) lie between \(a\) and \(1\) and must, therefore lie in the interval \(( a, b )\).

Since this is true for any interval \(( a, b )\) containing \(1\), it follows that \(1\) is the limit of the sequence.

Exercise 8.1.4

Show that \(\left\{\dfrac{2n+1}{n}\right\}\to 2\).

See Solution

Convergence of a series

Now that we know what it means for a sequence to converge to a limit \(L\), we can define what it means for a series to converge to a sum \(S\). But first, we need the concept of a sequence of partial sums.

Given the infinite series \(c_1 + c_2 + c_3 + \cdots\), define the sequence of partial sums as

\(\begin{eqnarray*} s_1 &=& c_1\\ s_2 &=& c_1 + c_2\\ s_3 &=& c_1 + c_2 + c_3\\ &\vdots&\\ s_n &=& c_1 + c_2 + c_3 + \cdots + c_n \end{eqnarray*}\)

Then the sequence of partial sums is \(s_1, s_2, s_3, \cdots, s_n, \cdots\)

To say that the series \(c_1 + c_2 + c_3 + \cdots\) has the sum \(L\) means that the sequence of partial sums \(s_1, s_2, s_3, \cdots, s_n, \cdots\) converges to the limit \(L\).

Let us show that \(1 + 0.1 + 0.01 + 0.001 + \cdots = \frac{10}{9}\).

Notice that the decimal expansion of \(\frac{10}{9}\) is \(1.1111\cdots\).

The sequence of partial sums is an increasing sequence

\( \begin{eqnarray*} 1,~1.1,~1.11,\cdots &=&\frac{1}{1},\frac{11}{10},\frac{111}{100},\cdots\\ &=&\frac{9}{9},\frac{99}{90},\frac{999}{900}\cdots\\ &=&\left(\frac{10}{9}-\frac{1}{9}\right),\left(\frac{100}{90}-\frac{1}{90}\right),\left(\frac{1000}{900}-\frac{1}{900}\right),\cdots\\ &=&\left(\frac{10}{9}-\frac{1}{9}\right),\left(\frac{10}{9}-\frac{1}{90}\right),\left(\frac{10}{9}-\frac{1}{900}\right),\cdots\\ &=&\left(\frac{10}{9}-\frac{1}{9}\left(\frac{1}{10}\right)^0\right),\left(\frac{10}{9}-\frac{1}{9}\left(\frac{1}{10}\right)^1\right),\left(\frac{10}{9}-\frac{1}{9}\left(\frac{1}{10}\right)^2\right),\cdots\\ &=&\left\{\frac{10}{9}-\frac{1}{9}\left(\frac{1}{10}\right)^{n-1}\right\} \end{eqnarray*} \)

Let \(( a, b )\) denote an interval containing the number \(\frac{10}{9}\). Then \(a < \frac{10}{9}\). Since we can find a value of \(n\) making the quantity \(\frac{1}{9}\left(\frac{1}{10}\right)^{n-1}\)as small as we please, pick \(n\) so that the quantity is smaller than \(\frac{10}{9}-a\).

Then it follows that \(a<\frac{10}{9}-\frac{1}{9}\left(\frac{1}{10}\right)^{n-1}\), which is the first term of a tail of the sequence.

Since the sequence is increasing, the interval \(( a, b )\) contains a tail of the sequence.

Since this will be true for any interval containing \(\frac{10}{9}\), that must be the limit of the sequence of partial sums.

Thus \(\frac{10}{9}\) is the sum of the series.

Exercise 8.1.5

Find a formula for the \(n\)-th partial sum of the sequence of partial sums of the sequence \(1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots\)

Prove that \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=2\)

See Solution

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