Calculus – Sequences and series – Sequences

Definition: A sequence \(\{a_n\}_{n=1}^{\infty}\) is an ordered list of numbers: \(a_1,a_2,a_3,\ldots\).

Examples:

1. \(1,2,3,4,5,\ldots\) is the sequence of positive integers \(\{a_n\}_{n=1}^{\infty}\) with \(a_n=n\) for \(n=1,2,3,\ldots\).


2. \(1,\frac{1}{4},\frac{1}{9},\frac{1}{16},\frac{1}{25},\ldots\) is the sequence \(\{a_n\}_{n=1}^{\infty}\) with \(a_n=\displaystyle\frac{1}{n^2}\) for \(n=1,2,3,\ldots\).


3. \(\frac{1}{2},\frac{2}{5},\frac{3}{10},\frac{4}{17},\frac{5}{26},\ldots\) is the sequence \(\{a_n\}_{n=1}^{\infty}\) with \(a_n=\displaystyle\frac{n}{n^2+1}\) for \(n=1,2,3,\ldots\).


4. \(1,-\frac{1}{2},\frac{1}{4},-\frac{1}{8},\frac{1}{16},-\frac{1}{32},\ldots\) is the sequence \(\{a_n\}_{n=0}^{\infty}\) with \(a_n=\displaystyle\frac{(-1)^n}{2^n}\) for \(n=0,1,2,\ldots\).


5. The sequence \(\{a_n\}_{n=1}^{\infty}\) defined by \(a_{n+2}=a_n+a_{n+1}\) with \(n=1,2,3,\ldots\) and \(a_1=a_2=1\) is called the Fibonacci sequence:
\[1,1,2,3,5,8,13,21,34,55,89,\ldots.\]

Definition: A sequence \(\{a_n\}_{n=1}^{\infty}\) is called convergent if \(\lim\limits_{n\to\infty}a_n\) exists. Otherwise, the sequence is called divergent.

Theorem: If \(\lim\limits_{x\to\infty}f(x)=L\) and \(f(n)=a_n\) for \(n=1,2,3,\ldots\), then \(\lim\limits_{n\to\infty}a_n=L\).

Theorem: If \(\lim\limits_{n\to\infty}|a_n|=0\), then \(\lim\limits_{n\to\infty}a_n=0\).

We can prove this by using the squeeze theorem since we have \(-|a_n|\leq a_n\leq|a_n|\) for all \(n\).

Theorem: If \(\lim\limits_{x\to\infty}f(x)=L\) and the function \(\) is continuous at \(L\), then \(\lim\limits_{n\to\infty}f(a_n)=f(L)\).

Definition: A sequence \(\{a_n\}_{n=1}^{\infty}\) is called increasing if \(a_n < a_{n+1}\) for all \(n\geq1\), that is \(a_1 < a_2 < a_3 <\cdots\).
It is called decreasing if \(a_n>a_{n+1}\) for all \(n\geq1\) that is \(a_1 > a_2 > a_3 >\cdots\).
A sequence is called monotonic if it is either increasing or decreasing.

In the case that \(a_n\leq a_{n+1}\) for all \(n\geq1\) the sequence is sometimes called not decreasing and if \(a_n\geq a_{n+1}\) for all \(n\) it is called not increasing.

Definition: A sequence \(\{a_n\}_{n=1}^{\infty}\) is called bounded above if there exists a number \(M\) such that \(a_n\leq M\) for all \(n\geq1\).
It is called bounded below if there exists a number \(m\) such that \(m\leq a_n\) for all \(n\geq1\).
A sequence is called bounded if it is both bounded above and bounded below.

Theorem: Every increasing sequence that is bounded above is convergent and every decreasing sequence that is bounded below is convergent.

Examples

Stewart §11.1, Example 14
Consider the sequence \(\{a_n\}_{n=1}^{\infty}\) defined by the recurrence relation \(a_1=2\) and \(a_{n+1}=\frac{1}{2}(a_n+6)\) for \(n=1,2,3,\ldots\).

Using the principle of mathematical induction we can prove that the sequence is increasing and bounded above.

Note that \(a_2=\frac{1}{2}(a_1+6)=\frac{1}{2}(2+6)=4>2=a_1\). Assume that \(a_{n+1} > a_n\) for a certain value of \(n\), then we have:

\[a_{n+2}=\frac{1}{2}(a_{n+1}+6) > \frac{1}{2}(a_n+6)=a_{n+1}.\]

This proves that the sequence is increasing.

Now, let us prove that the sequence is bounded above by \(6\). It is clear that \(a_1=2\leq6\). Assume that \(a_n\leq6\) for a certain value of \(n\), then we have:

\[a_{n+1}=\frac{1}{2}(a_n+6)\leq\frac{1}{2}(6+6)=6.\]

This proves that the sequence is bounded above.

Since \(\{a_n\}\) is an increasing sequence that is bounded above, the limit \(\lim\limits_{n\to\infty}a_n=L\) exists. Then we have:

\[\lim\limits_{n\to\infty}a_{n+1}=\lim\limits_{n\to\infty}\frac{1}{2}(a_n+6)=\frac{1}{2}\left(\lim\limits_{n\to\infty}a_n+6\right)=\frac{1}{2}\left(L+6\right).\]

Since \(\lim\limits_{n\to\infty}a_n=L\) it follows that \(\lim\limits_{n\to\infty}a_{n+1}=L\) as well. So we have: \(L=\frac{1}{2}(L+6)\). This implies that \(2L=L+6\) or equivalently \(L=6\).

Stewart §11.1, Exercise 79
Consider the sequence \(\left\{\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},\ldots\right\}\). Note that \(a_1=\sqrt{2}\) and \(a_{n+1}=\sqrt{2a_n}\) for \(n=1,2,3,\ldots\).

Then we have: \(a_2=\sqrt{2a_1}=\sqrt{2\sqrt{2}} > \sqrt{2}=a_1\). If we assume that \(a_{n+1} > a_n\) for a certain value of \(n\), then we have:

\[a_{n+2}=\sqrt{2a_{n+1}} > \sqrt{2a_n}=a_{n+1}.\]

This proves that the sequence is increasing.

Now, let us prove that the sequence is bounded above by \(2\). It is clear that \(a_1=\sqrt{2}\leq2\). Assume that \(a_n\leq2\) for a certain value of \(n\), then we have:

\[a_{n+1}=\sqrt{2a_n}\leq\sqrt{2\cdot2}=2.\]

This proves that the sequence is bounded above.

Since \(\{a_n\}\) is an increasing sequence that is bounded above, the limit \(\lim\limits_{n\to\infty}a_n=L\) exists. Then we have:

\[L=\sqrt{2L}\quad\Longrightarrow\quad L^2=2L\quad\Longrightarrow\quad L=0\;\;\text{or}\;\;L=2.\]

Since \(a_1=\sqrt{2}>0\) and the sequence is increasing, we conclude that \(L=2\).

Stewart §11.1, Exercise 80
Consider the sequence \(\left\{\sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}},\ldots\right\}\). Note that \(a_1=\sqrt{2}\) and \(a_{n+1}=\sqrt{2+a_n}\) for \(n=1,2,3,\ldots\).

Then we have: \(a_2=\sqrt{2+a_1}=\sqrt{2+\sqrt{2}} > \sqrt{2}=a_1\). If we assume that \(a_{n+1} > a_n\) for a certain value of \(n\), then we have:

\[a_{n+2}=\sqrt{2+a_{n+1}} > \sqrt{2+a_n}=a_{n+1}.\]

This proves that the sequence is increasing.

Now, let us prove that the sequence is bounded above by \(2\). It is clear that \(a_1=\sqrt{2}\leq2\). Assume that \(a_n\leq2\) for a certain value of \(n\), then we have:

\[a_{n+1}=\sqrt{2+a_n}\leq\sqrt{2+2}=2.\]

This proves that the sequence is bounded above.

Since \(\{a_n\}\) is an increasing sequence that is bounded above, the limit \(\lim\limits_{n\to\infty}a_n=L\) exists. Then we have:

\[L=\sqrt{2+L}\quad\Longrightarrow\quad L^2=2+L\quad\Longrightarrow\quad L=-1\;\;\text{or}\;\;L=2.\]

Since \(a_1=\sqrt{2}>0\) and the sequence is increasing, we conclude that \(L=2\).

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Last modified on March 15, 2021