7.7 The Translation Theorem

7.73   Theorem. Let $f$ be a real convergent sequence, say $f\to L$. If $f(n)\geq 0$ for all $n\in\mbox{{\bf N}}$, then $L\geq 0$.

Proof: I note that $L\in\mbox{{\bf R}}$, since if $f\to L$, then $\mbox{\rm Re}f\to\mbox{\rm Re}L$. Suppose, to get a contradiction, that $L<0$, (so $${-{L\over 2}>0}$$), and let $N_{f-\tilde L}$ be a precision function for the null sequence $f-\tilde L$. Let $${N=N_{f-\tilde L}\left(-{L\over 2}\right)}$$. Then $${\vert(f-\tilde L)(N)\vert\leq -{L\over 2}}$$, so $${\vert f(N)-L\vert\leq -{L\over 2}}$$, and hence $${f(N)<L-{L\over 2}={L\over 2}<0}$$. This contradicts the assumption that $f(N)\geq 0$ for all $n\in\mbox{{\bf N}}$. $\mid\!\mid\!\mid$

7.74   Exercise (Inequality theorem.) A Let $f,g$ be convergent real sequences. Suppose that $f(n)\leq g(n)$ for all $n\in\mbox{{\bf N}}$. Prove that $\lim f\leq\lim g$.

7.75   Exercise. A Prove the following assertion, or give an example to show that it is not true. Let $f,g$ be convergent real sequences. Suppose that $f(n)<g(n)$ for all $n\in\mbox{{\bf N}}$. Then $\lim f<\lim g$.

7.76   Definition (Translate of a sequence.) Let $f$ be a sequence and let $p\in\mbox{{\bf N}}$. Then the sequence $f_p\colon n\mapsto f(n+p)$ is called a translate of $f$.

7.77   Example. If $${f=\left\{ {1\over {2^2}},{1\over {3^2}},{1\over {4^2}},\cdots,{1\over {(n+2)^2}},\cdots\right\}}$$, then $${f_3=\left\{ {1\over {5^2}},{1\over {6^2}},{1\over {7^2}},\cdots,{1\over {(n+5)^2}},\cdots\right\}}$$. A translate of a sequence is a sequence obtained by ignoring the first few terms.

7.78   Theorem (Translation theorem.) If $\{f(n)\}$ is a convergent complex sequence, and $p\in\mbox{{\bf N}}$, then $\{f(n+p)\}$ converges, and $\lim\{f(n)\}=\lim\{f(n+p)\}$. Conversely, if $\{f(n+p)\}$ converges, then $\{f(n)\}$ converges to the same limit.

Proof: Let $f\to L$, let $f_p(n)=f(n+p)$ and let $N_{f-\tilde L}$ be a precision function for $f-\tilde L$. I claim $N_{f-\tilde L}$ is also a precision function for $f_p-\tilde L$. In fact, for all $n\in\mbox{{\bf N}}$, and all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$,

$$n\geq N_{f-\tilde L}(\varepsilon)\mbox{$\hspace{1ex}\Longrigh... ...x}\Longrightarrow\hspace{1ex}$}\vert f(n+p)-L\vert<\varepsilon.$$

Conversely, suppose

$$\{f_p(n)\}=\{f(n+p)\}\to L$$

and let $N_{f_p-\tilde L}$ be a precision function for $f_p-\tilde L$. Let $N(\varepsilon)=p + N_{f_p-\tilde L}(\varepsilon)$ for all $\varepsilon\in\mbox{${\mbox{{\bf R}}}^{+}$}$. I claim $N$ is a precision function for $N_{f-\tilde L}$. For all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} n>N(\varepsilon)&\mbox{$\Longrightarrow$}&n>p+N_{f_p-\tilde L}... ...arrow$}&\vert f(n)-L\vert<\varepsilon.\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

7.79   Example. Let the sequence $f$ be defined by

$$\begin{eqnarray*} f(0)&=&1, \\ f(n+1)&=&{1\over {1+f(n)}} \mbox{ for all } n\in\mbox{{\bf N}}. \end{eqnarray*}$$

Then

$$\begin{eqnarray*} f(1)&=&{1\over {1+1}}={1\over 2} \\ f(2)&=&{1\over {1+{1\over... ...1\over {1+{1\over {1+1}}}}}}={1\over {1+{2\over 3}}}={3\over 5}. \end{eqnarray*}$$

Suppose I knew that $f$ converged to a limit $L$. It is clear that $f(n)>0$ for all $n$, so $L$ must be $\geq 0$. By the translation theorem

$$L=\lim\{f(n+1)\}=\lim\left\{ {1\over {1+f(n)}}\right\}={1\over {1+\lim f(n)}}={1\over {1+L}}$$

so $L(1+L)=1$; i.e., $L^2+L-1=0$. Hence $${ L\in\left\{ {{-1+\sqrt{1+4}}\over 2},\; {{-1-\sqrt{1+4}}\over 2}\right\}}$$, and since $L\geq 0$, we conclude $${L={{\sqrt 5-1}\over 2}}$$. I've shown that the only thing that $f$ can possibly converge to is $${ {{\sqrt 5-1}\over 2}}$$. Now

$$0 < L < {3-1\over 2} = 1, \mbox{ so } \vert 1 - L \vert < 1.$$

Since $L = {1\over 1+L}$, we have for all $n\in\mbox{{\bf N}}$,

$$\begin{eqnarray*} \vert f(n+1)- L\vert &=& \left\vert {1\over 1+f(n)} - {1\over ... ...rt \over 1+L}\\ &=& L\vert L- f(n)\vert = L\vert f(n) - L\vert. \end{eqnarray*}$$

Hence

$$\begin{eqnarray*} \vert f(1) - L\vert & \leq & L\vert f(0) - L\vert = L\vert 1-L... ... \vert f(3) - L\vert & \leq & L\vert f(2) - L\vert \leq L^3,\\ \end{eqnarray*}$$

and by induction,

$$\vert f(n)-L\vert \leq L^n \mbox{ for all }n \in \mbox{{\bf Z}}_{\geq 1}.$$

By theorem 7.64 $\{L^n\}$ is a null sequence, and by the comparison theorem for null sequences, it follows that $\{f(n) - L\}$ is a null sequence. This completes the proof that $f\to L$. $\mid\!\mid\!\mid$

7.80   Exercise. Let

$$\begin{eqnarray*} f(0)&=&-2 \\ f(n+1)&=&{{f(n)^2+2}\over {2f(n)}} \mbox{ for all } n\in\mbox{{\bf N}}. \end{eqnarray*}$$

a)

Assume that $f$ converges, and determine the value of $\lim\{f(n)\}$.

b)

Calculate $f(1),f(2),f(3),f(4)$, using all of the accuracy of your calculator. Does the sequence appear to converge?

7.81   Entertainment. Show that the sequence $f$ defined in the previous exercise converges. We will prove this result in Example 7.97, but you can prove it now, using results you know.

7.82   Exercise. Let $g$ be the sequence defined by

$$\begin{eqnarray*} g(0) &=& 1,\\ g(1) &=& 1,\\ g(n+2) &=& {1+g(n+1) \over g(n)} \mbox{ for all }n \in\mbox{{\bf N}}. \end{eqnarray*}$$

a)

Assume that $g$ converges, and determine the value of $\lim\{g(n)\}$.

b)

Calculate $g(1),g(2),g(3),g(4),g(5),g(6)$, using all of the accuracy of your calculator. Does this sequence converge?

7.83   Theorem (Divergence test.) Let $f,g$ be complex sequences such that $g(n)\neq 0$ for all $n\in\mbox{{\bf N}}$. Suppose that $g\to 0$ and $f\to L$ where $L\neq 0$. Then $${{f\over g}}$$ diverges.

Proof: Suppose, to get a contradiction, that $${{f\over g}}$$ converges to a limit $M$. Then by the product theorem, $${g\cdot{f\over g}}$$ converges to $0\cdot M=0$; i.e., $f\to 0$. This contradicts our assumption that $f$ has a non-zero limit. $\mid\!\mid\!\mid$

7.84   Exercise. Prove the following assertion or give an example to show that it is not true: Let $f,g$ be complex sequences such that $g(n)\neq 0$ for all $n\in\mbox{{\bf N}}$, but $g\to 0$. Then $${{f\over g}}$$ diverges.

7.85   Example. Let $${f(n)=\left\{ {{n^3+3n}\over {n^2+1}}\right\}}$$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. Then

$$f(n)={{n^3\left(1+{3\over {n^2}}\right)}\over {n^2\left(1+{1\... ...r {n^2}}\right)}\over { {1\over n} \left(1+{1\over n}\right)}}.$$

Since

$$\lim\left\{\left(1+{3\over {n^2}}\right)\right\}_{n\geq 1}=1+0\neq 0,$$

and

$$\lim\left\{{1\over n}\left(1+{1\over n}\right)\right\}_{n\geq 1}=0\cdot(1+0)=0,$$

it follows that $f$ diverges.

7.86   Exercise. A Let $A,B,a,b$ be complex numbers such that $an+b\neq 0$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$. Discuss the convergence of $${\left\{ {{An+B}\over {an+b}}\right\}_{n\geq 1}}$$. Consider all possible choices for $A,B,a,b$.